Compare nested objects in JavaScript and return keys equality The 2019 Stack Overflow Developer Survey Results Are InJavaScript: Deep comparison recursively: Objects and propertiesGet the property of the difference between two objects in javascriptLength of a JavaScript objectWhat is the most efficient way to deep clone an object in JavaScript?How do I remove a property from a JavaScript object?How do I check if an array includes an object in JavaScript?Which equals operator (== vs ===) should be used in JavaScript comparisons?Compare two dates with JavaScriptHow do I test for an empty JavaScript object?How do I loop through or enumerate a JavaScript object?How do I correctly clone a JavaScript object?Checking if a key exists in a JavaScript object?

Ubuntu Server install with full GUI

Button changing its text & action. Good or terrible?

Keeping a retro style to sci-fi spaceships?

Why was M87 targeted for the Event Horizon Telescope instead of Sagittarius A*?

Can a flute soloist sit?

Are there any other methods to apply to solving simultaneous equations?

Why “相同意思的词” is called “同义词” instead of "同意词"?

Pokemon Turn Based battle (Python)

Star Trek - X-shaped Item on Regula/Orbital Office Starbases

Dropping list elements from nested list after evaluation

Is an up-to-date browser secure on an out-of-date OS?

Likelihood that a superbug or lethal virus could come from a landfill

Cooking pasta in a water boiler

What to do when moving next to a bird sanctuary with a loosely-domesticated cat?

Why does the nucleus not repel itself?

A word that means fill it to the required quantity

What is this sharp, curved notch on my knife for?

Relationship between Gromov-Witten and Taubes' Gromov invariant

What is this business jet?

Loose spokes after only a few rides

APIPA and LAN Broadcast Domain

Is it possible for absolutely everyone to attain enlightenment?

Correct punctuation for showing a character's confusion

Can I have a signal generator on while it's not connected?



Compare nested objects in JavaScript and return keys equality



The 2019 Stack Overflow Developer Survey Results Are InJavaScript: Deep comparison recursively: Objects and propertiesGet the property of the difference between two objects in javascriptLength of a JavaScript objectWhat is the most efficient way to deep clone an object in JavaScript?How do I remove a property from a JavaScript object?How do I check if an array includes an object in JavaScript?Which equals operator (== vs ===) should be used in JavaScript comparisons?Compare two dates with JavaScriptHow do I test for an empty JavaScript object?How do I loop through or enumerate a JavaScript object?How do I correctly clone a JavaScript object?Checking if a key exists in a JavaScript object?



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








7















I have two nested objects obj1 and obj2 and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag



So for a given obj1 like



obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"




and the obj2 like



obj2 = 
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new"



it should return



equality = 
prop1: false,
prop2: true,
prop3 :
prop4: true,
prop5: false
,
prop6: false



If an object has a new property, like obj2.prop6, then the equality will be equality.prop6 = false.



For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties










share|improve this question



















  • 1





    Will both objects always have exact match properties?

    – holydragon
    2 days ago











  • good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

    – loretoparisi
    2 days ago


















7















I have two nested objects obj1 and obj2 and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag



So for a given obj1 like



obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"




and the obj2 like



obj2 = 
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new"



it should return



equality = 
prop1: false,
prop2: true,
prop3 :
prop4: true,
prop5: false
,
prop6: false



If an object has a new property, like obj2.prop6, then the equality will be equality.prop6 = false.



For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties










share|improve this question



















  • 1





    Will both objects always have exact match properties?

    – holydragon
    2 days ago











  • good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

    – loretoparisi
    2 days ago














7












7








7


1






I have two nested objects obj1 and obj2 and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag



So for a given obj1 like



obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"




and the obj2 like



obj2 = 
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new"



it should return



equality = 
prop1: false,
prop2: true,
prop3 :
prop4: true,
prop5: false
,
prop6: false



If an object has a new property, like obj2.prop6, then the equality will be equality.prop6 = false.



For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties










share|improve this question
















I have two nested objects obj1 and obj2 and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag



So for a given obj1 like



obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"




and the obj2 like



obj2 = 
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new"



it should return



equality = 
prop1: false,
prop2: true,
prop3 :
prop4: true,
prop5: false
,
prop6: false



If an object has a new property, like obj2.prop6, then the equality will be equality.prop6 = false.



For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties







javascript






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 days ago







loretoparisi

















asked 2 days ago









loretoparisiloretoparisi

8,08154973




8,08154973







  • 1





    Will both objects always have exact match properties?

    – holydragon
    2 days ago











  • good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

    – loretoparisi
    2 days ago













  • 1





    Will both objects always have exact match properties?

    – holydragon
    2 days ago











  • good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

    – loretoparisi
    2 days ago








1




1





Will both objects always have exact match properties?

– holydragon
2 days ago





Will both objects always have exact match properties?

– holydragon
2 days ago













good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

– loretoparisi
2 days ago






good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

– loretoparisi
2 days ago













4 Answers
4






active

oldest

votes


















1














You could use reduce to build new object and another get method to get nested props from other object by string and compare it to current prop value in first object.






const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" 
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"

function get(obj, path)
return path.split('.').reduce((r, e) =>
if (!r) return r
else return r[e] , obj)


function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )


const result = compare(obj1, obj2);
console.log(result)





To compare all properties of both objects you could create extra function that will perform loop by both objects.






const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"

function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);


function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;


function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;

if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);

else
r[e] = value;

return r;
,
o ? o :
);


function compare(a, b)
const o = build(a, b);
return build(b, a, o);


const result = compare(obj1, obj2);
console.log(result)








share|improve this answer




















  • 1





    I thinks this is the best solution since it supports all recent version of ECMAScript.

    – loretoparisi
    2 days ago











  • Just a thing, supposed that the property is a void object like , it does not make the comparison.

    – loretoparisi
    2 days ago






  • 1





    In that case you could use this jsfiddle.net/vqwn3zLf

    – Nenad Vracar
    2 days ago











  • Ok, now it handles a nested lonely prop11= object, but if you have like the same prop12= in both, you will get the as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

    – loretoparisi
    2 days ago







  • 1





    @loretoparisi Try this jsfiddle.net/r0y8nd3q/1

    – Nenad Vracar
    2 days ago


















5














You could iterate all keys and check the nested objects if both values are objects.






const isObject = v => v && typeof v === 'object';

function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));


var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;

console.log(getDifference(obj1, obj2));








share|improve this answer

























  • Thanks! This solution works in every condition even if the property is a void object like prop7 = . Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

    – loretoparisi
    2 days ago







  • 1





    without spread not Set.

    – Nina Scholz
    2 days ago











  • maybe like Object.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))

    – loretoparisi
    2 days ago






  • 1





    w h a t e v e r ...

    – Nina Scholz
    2 days ago


















2














Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value || check is added.






const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";

const isObject = val => val && typeof val === 'object'; // required for "null" comparison

function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both

for (let key in merged)

return equality;


console.log(compare(obj1, obj2))








share|improve this answer

























  • Thanks, just updated the code, assuming you can have new properties as well.

    – loretoparisi
    2 days ago











  • what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

    – AZ_
    2 days ago











  • @loretoparisi updated

    – adiga
    2 days ago











  • @adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

    – AZ_
    2 days ago












  • @AZ_ updated. Not sure if it will fail for any scenario

    – adiga
    2 days ago


















1














A recursive example,






var obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,


var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,


var result = ;

function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);

else
obj_[k] = (obj1[k] === obj2[k]);





if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;


compare(obj1, obj2, result);

console.log(result);








share|improve this answer

























  • Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a in the results for properties of type object that belong to both objects and that are .

    – loretoparisi
    2 days ago







  • 1





    @loretoparisi I made a change, try now.

    – Shoyeb Sheikh
    2 days ago











Your Answer






StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55591096%2fcompare-nested-objects-in-javascript-and-return-keys-equality%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














You could use reduce to build new object and another get method to get nested props from other object by string and compare it to current prop value in first object.






const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" 
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"

function get(obj, path)
return path.split('.').reduce((r, e) =>
if (!r) return r
else return r[e] , obj)


function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )


const result = compare(obj1, obj2);
console.log(result)





To compare all properties of both objects you could create extra function that will perform loop by both objects.






const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"

function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);


function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;


function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;

if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);

else
r[e] = value;

return r;
,
o ? o :
);


function compare(a, b)
const o = build(a, b);
return build(b, a, o);


const result = compare(obj1, obj2);
console.log(result)








share|improve this answer




















  • 1





    I thinks this is the best solution since it supports all recent version of ECMAScript.

    – loretoparisi
    2 days ago











  • Just a thing, supposed that the property is a void object like , it does not make the comparison.

    – loretoparisi
    2 days ago






  • 1





    In that case you could use this jsfiddle.net/vqwn3zLf

    – Nenad Vracar
    2 days ago











  • Ok, now it handles a nested lonely prop11= object, but if you have like the same prop12= in both, you will get the as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

    – loretoparisi
    2 days ago







  • 1





    @loretoparisi Try this jsfiddle.net/r0y8nd3q/1

    – Nenad Vracar
    2 days ago















1














You could use reduce to build new object and another get method to get nested props from other object by string and compare it to current prop value in first object.






const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" 
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"

function get(obj, path)
return path.split('.').reduce((r, e) =>
if (!r) return r
else return r[e] , obj)


function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )


const result = compare(obj1, obj2);
console.log(result)





To compare all properties of both objects you could create extra function that will perform loop by both objects.






const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"

function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);


function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;


function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;

if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);

else
r[e] = value;

return r;
,
o ? o :
);


function compare(a, b)
const o = build(a, b);
return build(b, a, o);


const result = compare(obj1, obj2);
console.log(result)








share|improve this answer




















  • 1





    I thinks this is the best solution since it supports all recent version of ECMAScript.

    – loretoparisi
    2 days ago











  • Just a thing, supposed that the property is a void object like , it does not make the comparison.

    – loretoparisi
    2 days ago






  • 1





    In that case you could use this jsfiddle.net/vqwn3zLf

    – Nenad Vracar
    2 days ago











  • Ok, now it handles a nested lonely prop11= object, but if you have like the same prop12= in both, you will get the as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

    – loretoparisi
    2 days ago







  • 1





    @loretoparisi Try this jsfiddle.net/r0y8nd3q/1

    – Nenad Vracar
    2 days ago













1












1








1







You could use reduce to build new object and another get method to get nested props from other object by string and compare it to current prop value in first object.






const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" 
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"

function get(obj, path)
return path.split('.').reduce((r, e) =>
if (!r) return r
else return r[e] , obj)


function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )


const result = compare(obj1, obj2);
console.log(result)





To compare all properties of both objects you could create extra function that will perform loop by both objects.






const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"

function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);


function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;


function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;

if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);

else
r[e] = value;

return r;
,
o ? o :
);


function compare(a, b)
const o = build(a, b);
return build(b, a, o);


const result = compare(obj1, obj2);
console.log(result)








share|improve this answer















You could use reduce to build new object and another get method to get nested props from other object by string and compare it to current prop value in first object.






const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" 
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"

function get(obj, path)
return path.split('.').reduce((r, e) =>
if (!r) return r
else return r[e] , obj)


function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )


const result = compare(obj1, obj2);
console.log(result)





To compare all properties of both objects you could create extra function that will perform loop by both objects.






const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"

function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);


function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;


function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;

if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);

else
r[e] = value;

return r;
,
o ? o :
);


function compare(a, b)
const o = build(a, b);
return build(b, a, o);


const result = compare(obj1, obj2);
console.log(result)








const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" 
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"

function get(obj, path)
return path.split('.').reduce((r, e) =>
if (!r) return r
else return r[e] , obj)


function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )


const result = compare(obj1, obj2);
console.log(result)





const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" 
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"

function get(obj, path)
return path.split('.').reduce((r, e) =>
if (!r) return r
else return r[e] , obj)


function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )


const result = compare(obj1, obj2);
console.log(result)





const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"

function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);


function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;


function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;

if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);

else
r[e] = value;

return r;
,
o ? o :
);


function compare(a, b)
const o = build(a, b);
return build(b, a, o);


const result = compare(obj1, obj2);
console.log(result)





const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"

function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);


function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;


function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;

if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);

else
r[e] = value;

return r;
,
o ? o :
);


function compare(a, b)
const o = build(a, b);
return build(b, a, o);


const result = compare(obj1, obj2);
console.log(result)






share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























answered 2 days ago









Nenad VracarNenad Vracar

73.4k126085




73.4k126085







  • 1





    I thinks this is the best solution since it supports all recent version of ECMAScript.

    – loretoparisi
    2 days ago











  • Just a thing, supposed that the property is a void object like , it does not make the comparison.

    – loretoparisi
    2 days ago






  • 1





    In that case you could use this jsfiddle.net/vqwn3zLf

    – Nenad Vracar
    2 days ago











  • Ok, now it handles a nested lonely prop11= object, but if you have like the same prop12= in both, you will get the as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

    – loretoparisi
    2 days ago







  • 1





    @loretoparisi Try this jsfiddle.net/r0y8nd3q/1

    – Nenad Vracar
    2 days ago












  • 1





    I thinks this is the best solution since it supports all recent version of ECMAScript.

    – loretoparisi
    2 days ago











  • Just a thing, supposed that the property is a void object like , it does not make the comparison.

    – loretoparisi
    2 days ago






  • 1





    In that case you could use this jsfiddle.net/vqwn3zLf

    – Nenad Vracar
    2 days ago











  • Ok, now it handles a nested lonely prop11= object, but if you have like the same prop12= in both, you will get the as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

    – loretoparisi
    2 days ago







  • 1





    @loretoparisi Try this jsfiddle.net/r0y8nd3q/1

    – Nenad Vracar
    2 days ago







1




1





I thinks this is the best solution since it supports all recent version of ECMAScript.

– loretoparisi
2 days ago





I thinks this is the best solution since it supports all recent version of ECMAScript.

– loretoparisi
2 days ago













Just a thing, supposed that the property is a void object like , it does not make the comparison.

– loretoparisi
2 days ago





Just a thing, supposed that the property is a void object like , it does not make the comparison.

– loretoparisi
2 days ago




1




1





In that case you could use this jsfiddle.net/vqwn3zLf

– Nenad Vracar
2 days ago





In that case you could use this jsfiddle.net/vqwn3zLf

– Nenad Vracar
2 days ago













Ok, now it handles a nested lonely prop11= object, but if you have like the same prop12= in both, you will get the as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

– loretoparisi
2 days ago






Ok, now it handles a nested lonely prop11= object, but if you have like the same prop12= in both, you will get the as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

– loretoparisi
2 days ago





1




1





@loretoparisi Try this jsfiddle.net/r0y8nd3q/1

– Nenad Vracar
2 days ago





@loretoparisi Try this jsfiddle.net/r0y8nd3q/1

– Nenad Vracar
2 days ago













5














You could iterate all keys and check the nested objects if both values are objects.






const isObject = v => v && typeof v === 'object';

function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));


var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;

console.log(getDifference(obj1, obj2));








share|improve this answer

























  • Thanks! This solution works in every condition even if the property is a void object like prop7 = . Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

    – loretoparisi
    2 days ago







  • 1





    without spread not Set.

    – Nina Scholz
    2 days ago











  • maybe like Object.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))

    – loretoparisi
    2 days ago






  • 1





    w h a t e v e r ...

    – Nina Scholz
    2 days ago















5














You could iterate all keys and check the nested objects if both values are objects.






const isObject = v => v && typeof v === 'object';

function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));


var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;

console.log(getDifference(obj1, obj2));








share|improve this answer

























  • Thanks! This solution works in every condition even if the property is a void object like prop7 = . Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

    – loretoparisi
    2 days ago







  • 1





    without spread not Set.

    – Nina Scholz
    2 days ago











  • maybe like Object.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))

    – loretoparisi
    2 days ago






  • 1





    w h a t e v e r ...

    – Nina Scholz
    2 days ago













5












5








5







You could iterate all keys and check the nested objects if both values are objects.






const isObject = v => v && typeof v === 'object';

function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));


var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;

console.log(getDifference(obj1, obj2));








share|improve this answer















You could iterate all keys and check the nested objects if both values are objects.






const isObject = v => v && typeof v === 'object';

function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));


var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;

console.log(getDifference(obj1, obj2));








const isObject = v => v && typeof v === 'object';

function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));


var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;

console.log(getDifference(obj1, obj2));





const isObject = v => v && typeof v === 'object';

function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));


var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;

console.log(getDifference(obj1, obj2));






share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























answered 2 days ago









Nina ScholzNina Scholz

197k15110179




197k15110179












  • Thanks! This solution works in every condition even if the property is a void object like prop7 = . Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

    – loretoparisi
    2 days ago







  • 1





    without spread not Set.

    – Nina Scholz
    2 days ago











  • maybe like Object.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))

    – loretoparisi
    2 days ago






  • 1





    w h a t e v e r ...

    – Nina Scholz
    2 days ago

















  • Thanks! This solution works in every condition even if the property is a void object like prop7 = . Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

    – loretoparisi
    2 days ago







  • 1





    without spread not Set.

    – Nina Scholz
    2 days ago











  • maybe like Object.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))

    – loretoparisi
    2 days ago






  • 1





    w h a t e v e r ...

    – Nina Scholz
    2 days ago
















Thanks! This solution works in every condition even if the property is a void object like prop7 = . Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

– loretoparisi
2 days ago






Thanks! This solution works in every condition even if the property is a void object like prop7 = . Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

– loretoparisi
2 days ago





1




1





without spread not Set.

– Nina Scholz
2 days ago





without spread not Set.

– Nina Scholz
2 days ago













maybe like Object.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))

– loretoparisi
2 days ago





maybe like Object.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))

– loretoparisi
2 days ago




1




1





w h a t e v e r ...

– Nina Scholz
2 days ago





w h a t e v e r ...

– Nina Scholz
2 days ago











2














Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value || check is added.






const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";

const isObject = val => val && typeof val === 'object'; // required for "null" comparison

function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both

for (let key in merged)

return equality;


console.log(compare(obj1, obj2))








share|improve this answer

























  • Thanks, just updated the code, assuming you can have new properties as well.

    – loretoparisi
    2 days ago











  • what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

    – AZ_
    2 days ago











  • @loretoparisi updated

    – adiga
    2 days ago











  • @adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

    – AZ_
    2 days ago












  • @AZ_ updated. Not sure if it will fail for any scenario

    – adiga
    2 days ago















2














Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value || check is added.






const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";

const isObject = val => val && typeof val === 'object'; // required for "null" comparison

function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both

for (let key in merged)

return equality;


console.log(compare(obj1, obj2))








share|improve this answer

























  • Thanks, just updated the code, assuming you can have new properties as well.

    – loretoparisi
    2 days ago











  • what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

    – AZ_
    2 days ago











  • @loretoparisi updated

    – adiga
    2 days ago











  • @adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

    – AZ_
    2 days ago












  • @AZ_ updated. Not sure if it will fail for any scenario

    – adiga
    2 days ago













2












2








2







Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value || check is added.






const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";

const isObject = val => val && typeof val === 'object'; // required for "null" comparison

function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both

for (let key in merged)

return equality;


console.log(compare(obj1, obj2))








share|improve this answer















Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value || check is added.






const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";

const isObject = val => val && typeof val === 'object'; // required for "null" comparison

function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both

for (let key in merged)

return equality;


console.log(compare(obj1, obj2))








const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";

const isObject = val => val && typeof val === 'object'; // required for "null" comparison

function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both

for (let key in merged)

return equality;


console.log(compare(obj1, obj2))





const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";

const isObject = val => val && typeof val === 'object'; // required for "null" comparison

function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both

for (let key in merged)

return equality;


console.log(compare(obj1, obj2))






share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























answered 2 days ago









adigaadiga

12.4k62745




12.4k62745












  • Thanks, just updated the code, assuming you can have new properties as well.

    – loretoparisi
    2 days ago











  • what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

    – AZ_
    2 days ago











  • @loretoparisi updated

    – adiga
    2 days ago











  • @adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

    – AZ_
    2 days ago












  • @AZ_ updated. Not sure if it will fail for any scenario

    – adiga
    2 days ago

















  • Thanks, just updated the code, assuming you can have new properties as well.

    – loretoparisi
    2 days ago











  • what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

    – AZ_
    2 days ago











  • @loretoparisi updated

    – adiga
    2 days ago











  • @adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

    – AZ_
    2 days ago












  • @AZ_ updated. Not sure if it will fail for any scenario

    – adiga
    2 days ago
















Thanks, just updated the code, assuming you can have new properties as well.

– loretoparisi
2 days ago





Thanks, just updated the code, assuming you can have new properties as well.

– loretoparisi
2 days ago













what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

– AZ_
2 days ago





what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

– AZ_
2 days ago













@loretoparisi updated

– adiga
2 days ago





@loretoparisi updated

– adiga
2 days ago













@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

– AZ_
2 days ago






@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

– AZ_
2 days ago














@AZ_ updated. Not sure if it will fail for any scenario

– adiga
2 days ago





@AZ_ updated. Not sure if it will fail for any scenario

– adiga
2 days ago











1














A recursive example,






var obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,


var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,


var result = ;

function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);

else
obj_[k] = (obj1[k] === obj2[k]);





if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;


compare(obj1, obj2, result);

console.log(result);








share|improve this answer

























  • Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a in the results for properties of type object that belong to both objects and that are .

    – loretoparisi
    2 days ago







  • 1





    @loretoparisi I made a change, try now.

    – Shoyeb Sheikh
    2 days ago















1














A recursive example,






var obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,


var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,


var result = ;

function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);

else
obj_[k] = (obj1[k] === obj2[k]);





if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;


compare(obj1, obj2, result);

console.log(result);








share|improve this answer

























  • Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a in the results for properties of type object that belong to both objects and that are .

    – loretoparisi
    2 days ago







  • 1





    @loretoparisi I made a change, try now.

    – Shoyeb Sheikh
    2 days ago













1












1








1







A recursive example,






var obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,


var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,


var result = ;

function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);

else
obj_[k] = (obj1[k] === obj2[k]);





if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;


compare(obj1, obj2, result);

console.log(result);








share|improve this answer















A recursive example,






var obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,


var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,


var result = ;

function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);

else
obj_[k] = (obj1[k] === obj2[k]);





if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;


compare(obj1, obj2, result);

console.log(result);








var obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,


var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,


var result = ;

function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);

else
obj_[k] = (obj1[k] === obj2[k]);





if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;


compare(obj1, obj2, result);

console.log(result);





var obj1 = 
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,


var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,


var result = ;

function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);

else
obj_[k] = (obj1[k] === obj2[k]);





if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;


compare(obj1, obj2, result);

console.log(result);






share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























answered 2 days ago









Shoyeb SheikhShoyeb Sheikh

777211




777211












  • Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a in the results for properties of type object that belong to both objects and that are .

    – loretoparisi
    2 days ago







  • 1





    @loretoparisi I made a change, try now.

    – Shoyeb Sheikh
    2 days ago

















  • Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a in the results for properties of type object that belong to both objects and that are .

    – loretoparisi
    2 days ago







  • 1





    @loretoparisi I made a change, try now.

    – Shoyeb Sheikh
    2 days ago
















Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a in the results for properties of type object that belong to both objects and that are .

– loretoparisi
2 days ago






Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a in the results for properties of type object that belong to both objects and that are .

– loretoparisi
2 days ago





1




1





@loretoparisi I made a change, try now.

– Shoyeb Sheikh
2 days ago





@loretoparisi I made a change, try now.

– Shoyeb Sheikh
2 days ago

















draft saved

draft discarded
















































Thanks for contributing an answer to Stack Overflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55591096%2fcompare-nested-objects-in-javascript-and-return-keys-equality%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Tamil (spriik) Luke uk diar | Nawigatjuun

Align equal signs while including text over equalitiesAMS align: left aligned text/math plus multicolumn alignmentMultiple alignmentsAligning equations in multiple placesNumbering and aligning an equation with multiple columnsHow to align one equation with another multline equationUsing \ in environments inside the begintabularxNumber equations and preserving alignment of equal signsHow can I align equations to the left and to the right?Double equation alignment problem within align enviromentAligned within align: Why are they right-aligned?

Where does the image of a data connector as a sharp metal spike originate from?Where does the concept of infected people turning into zombies only after death originate from?Where does the motif of a reanimated human head originate?Where did the notion that Dragons could speak originate?Where does the archetypal image of the 'Grey' alien come from?Where did the suffix '-Man' originate?Where does the notion of being injured or killed by an illusion originate?Where did the term “sophont” originate?Where does the trope of magic spells being driven by advanced technology originate from?Where did the term “the living impaired” originate?