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Regex in IF condition in awk

Regex for matchng anything between parenthesisawk adds extra comma at several placesUsing multiple awk commands within single lineBest way to iterate through the following awk commandIf column matches another file, print every line with match (awk/grep)How do I form a new string from a parsed CSV line?Print text before and after match, from a specific beginning and to an ending stringsearch column 2 in csv file for value, if value, then insert “invalid” and shift cells rightAwk for merging multiple files with common columnUsing awk to find matches and extract characters from BEFORE each match - help!

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?

Required regex condition: [[ "$1" =~ [A-Za-z] ]]

BEGIN FS = ";"; counter=0

 
 if ((length($1) != 10 && length($1) != 12))
 
 counter++
 print counter, $1;
 if ($counter -gt 2)
 print "Invalid input file";
 exit;
 
 

I am getting error if I use the same condition which I have posted. How to add the condition?

awk regular-expression

share|improve this question

edited Apr 15 at 13:05

muru

44.8k55 gold badges111111 silver badges184184 bronze badges

asked Apr 15 at 12:20

LaxmanLaxman

2544 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage
  
  – steeldriver
  Apr 15 at 12:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just [A-Za-z]. What are you adding to your awk?
  
  – terdon♦
  Apr 15 at 12:34
  

  
  
  
  

add a comment
 | 

2

I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?

Required regex condition: [[ "$1" =~ [A-Za-z] ]]

BEGIN FS = ";"; counter=0

 
 if ((length($1) != 10 && length($1) != 12))
 
 counter++
 print counter, $1;
 if ($counter -gt 2)
 print "Invalid input file";
 exit;
 
 

I am getting error if I use the same condition which I have posted. How to add the condition?

awk regular-expression

share|improve this question

edited Apr 15 at 13:05

muru

44.8k55 gold badges111111 silver badges184184 bronze badges

asked Apr 15 at 12:20

LaxmanLaxman

2544 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage
  
  – steeldriver
  Apr 15 at 12:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just [A-Za-z]. What are you adding to your awk?
  
  – terdon♦
  Apr 15 at 12:34
  

  
  
  
  

add a comment
 | 

I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?

Required regex condition: [[ "$1" =~ [A-Za-z] ]]

BEGIN FS = ";"; counter=0

 
 if ((length($1) != 10 && length($1) != 12))
 
 counter++
 print counter, $1;
 if ($counter -gt 2)
 print "Invalid input file";
 exit;
 
 

I am getting error if I use the same condition which I have posted. How to add the condition?

awk regular-expression

share|improve this question

edited Apr 15 at 13:05

muru

44.8k55 gold badges111111 silver badges184184 bronze badges

asked Apr 15 at 12:20

LaxmanLaxman

2544 bronze badges

BEGIN FS = ";"; counter=0

 
 if ((length($1) != 10 && length($1) != 12))
 
 counter++
 print counter, $1;
 if ($counter -gt 2)
 print "Invalid input file";
 exit;
 
 

share|improve this question

edited Apr 15 at 13:05

muru

44.8k55 gold badges111111 silver badges184184 bronze badges

asked Apr 15 at 12:20

LaxmanLaxman

2544 bronze badges

share|improve this question

edited Apr 15 at 13:05

muru

44.8k55 gold badges111111 silver badges184184 bronze badges

asked Apr 15 at 12:20

LaxmanLaxman

2544 bronze badges

edited Apr 15 at 13:05

muru

44.8k55 gold badges111111 silver badges184184 bronze badges

edited Apr 15 at 13:05

muru

44.8k55 gold badges111111 silver badges184184 bronze badges

asked Apr 15 at 12:20

LaxmanLaxman

2544 bronze badges

asked Apr 15 at 12:20

LaxmanLaxman

2544 bronze badges

1 Answer
1

active

oldest

votes

8

You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.

BEGIN FS = ";"; counter=0

 
 if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)
 
 counter++
 print counter, $1;
 if (counter > 2)
 print "Invalid input file";
 exit;
 
 

Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.

share|improve this answer

edited Apr 15 at 16:58

answered Apr 15 at 12:36

terdon♦terdon

143k3535 gold badges295295 silver badges472472 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes. I was using =~ format.
  
  – Laxman
  Apr 15 at 12:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why couldnt you move the condition to be the pattern? length($1) != 10 && ... && $1 ~ /[A-Za-z]/ counter++ ... seems more « awkish » to me, avoiding some seriously nested braces
  
  – D. Ben Knoble
  Apr 15 at 16:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!
  
  – terdon♦
  Apr 15 at 17:00
  

  
  
  
  

add a comment
 | 

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8

You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.

BEGIN FS = ";"; counter=0

 
 if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)
 
 counter++
 print counter, $1;
 if (counter > 2)
 print "Invalid input file";
 exit;
 
 

Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.

share|improve this answer

edited Apr 15 at 16:58

answered Apr 15 at 12:36

terdon♦terdon

143k3535 gold badges295295 silver badges472472 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes. I was using =~ format.
  
  – Laxman
  Apr 15 at 12:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why couldnt you move the condition to be the pattern? length($1) != 10 && ... && $1 ~ /[A-Za-z]/ counter++ ... seems more « awkish » to me, avoiding some seriously nested braces
  
  – D. Ben Knoble
  Apr 15 at 16:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!
  
  – terdon♦
  Apr 15 at 17:00
  

  
  
  
  

add a comment
 | 

8

You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.

BEGIN FS = ";"; counter=0

 
 if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)
 
 counter++
 print counter, $1;
 if (counter > 2)
 print "Invalid input file";
 exit;
 
 

Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.

share|improve this answer

edited Apr 15 at 16:58

answered Apr 15 at 12:36

terdon♦terdon

143k3535 gold badges295295 silver badges472472 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes. I was using =~ format.
  
  – Laxman
  Apr 15 at 12:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why couldnt you move the condition to be the pattern? length($1) != 10 && ... && $1 ~ /[A-Za-z]/ counter++ ... seems more « awkish » to me, avoiding some seriously nested braces
  
  – D. Ben Knoble
  Apr 15 at 16:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!
  
  – terdon♦
  Apr 15 at 17:00
  

  
  
  
  

add a comment
 | 

You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.

BEGIN FS = ";"; counter=0

 
 if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)
 
 counter++
 print counter, $1;
 if (counter > 2)
 print "Invalid input file";
 exit;
 
 

Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.

share|improve this answer

edited Apr 15 at 16:58

answered Apr 15 at 12:36

terdon♦terdon

143k3535 gold badges295295 silver badges472472 bronze badges

BEGIN FS = ";"; counter=0

 
 if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)
 
 counter++
 print counter, $1;
 if (counter > 2)
 print "Invalid input file";
 exit;
 
 

share|improve this answer

edited Apr 15 at 16:58

answered Apr 15 at 12:36

terdon♦terdon

143k3535 gold badges295295 silver badges472472 bronze badges

answered Apr 15 at 12:36

terdon♦terdon

143k3535 gold badges295295 silver badges472472 bronze badges

answered Apr 15 at 12:36

terdon♦terdon

143k3535 gold badges295295 silver badges472472 bronze badges