Bsrgvty

I wrote this script :

#!/bin/bash
while [ true ] 
do
 currentoutput="$(lsusb)"
 if [ "$currentoutput" != "$lastoutput" ]
 then
 echo "" date and Time >> test.log
 date +%x_r >> test.log
 lastoutput="$(lsusb)"
 lsusb >> test.log
 fi
 sleep 5
done

I'm a newbie trying to learn fast and I got a question about the variable’s quotation marks.

Put a variable between $(), I get it, but why are the quotation marks needed even in the if statement? Is it to make a nested command?

Quotation marks prevent "word splitting". That is: breaking down variables into multiple items at whitespace characters (or to be more exact, at spaces, tabs, and newlines as defined in the value of the default $IFS shell variable).

For example,

$ var="one two"
$ howmany() echo $#; 
$ howmany $var
2
$ howmany "$var"
1

Here we define the howmany function which just lets us know how many positional parameters are given. As you can see, there are two items being passed to the variable, and with the quotes the text in the variable is treated as one unit.

This is important for accurate passing of information. For example, if the variable contains path to file, and the filename contains spaces anywhere in the path, the command you are trying to run may fail or give inaccurate results. If we were trying to create a file with the $var variable, touch $var would create two files, but touch "$var" just one.

Same goes for your [ "$currentoutput" != "$lastoutput" ] part. This particular test performs a comparison on two strings. When the test runs, the [ command would need to see 3 arguments - a text string, the != operator, and another text string. Keeping double quotes prevents word splitting, and the [ command sees exactly those 3 arguments. Now what happens if variables are unquoted ?

$ var="hello world"
$ foo="hi world"
$ [ $var != $foo ]
bash: [: too many arguments
$ 

Here, word splitting occurs, and instead [ sees two strings hello and world followed by !=, followed by two other strings hi world. Key point is that without double quotes, the contents of variables are understood as separate units rather than one whole item.

Assigning command substitution doesn't require double quotes as in

var=$( df )

where you have the df command's output saved to var. However, it is a good habit to always double quote variables and command substitution $(...) unless you do in fact want the output to be treated as separate items.

On a side note, the

while [ true ]

part can be

while true

[ is a command which evaluates its arguments, and [ whatever ] is always true regardless of what is inside. By contrast, while true uses the command true which always returns success exit status (and that's exactly what while loop needs). The difference is a bit more clarity and less testing performed. Alternatively, you could also use : instead of true

The double quotes in echo "" date and Time part could probably be removed. They merely insert an empty string and an add extra space to the output. If that's desired, feel free to keep them there, but there's no particular functional value in this case.

lsusb >> test.log

This part could probably be replaced with echo "$currentoutput" >> test.log. There's no reason to run lsusb again after it has been run already in currentoutput=$(lsusb). In cases where trailing newlines
have to be preserved in the output - one could see the value in running a command multiple times, but in case of lsusb there's no need for that. The less external commands you call, the better, because every call to a non-built-in command incurs costs in CPU, memory usage, and execution time ( even though the commands are probably pre-loaded from memory).

See also:

• When is double-quoting necessary?


• What is the point of the bash null operator : (colon)?


• Security implications of forgetting to quote a variable in bash/POSIX shells

In currentoutput="$(lsusb)" lsusb is not a variable, it is a command. What this statement does, it executes lsusb command and assigns its output to currentoutput variable.

Older syntax for this was

currentoutput=`lsusb`

you can find it in many examples and scripts

To answer the other part of your question, if [ ] is just how syntax for if is defined in bash. See more in https://www.tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html

The following runs the external command command and returns its output.

"$(command)"

Without the brackets/parentheses, this would look for a variable instead of running a command:

"$variable"

As for the difference between $variable and "$variable", this becomes relevant when $variable contains spaces. When using "$variable", the entire variable contents will be inserted into a single string even if the contents include spaces. When using $variable the contents of the variable may be expanded into an argument list of multiple arguments.

To be contrarian - bash recommends you use the [[ ... ]] over the [ ... ] construct to avoid having to quote variables in the test and so the associated problems of word splitting that the others have pointed out.

[ is provided in bash for POSIX compatibility with scripts meant to run under #!/bin/sh or those being ported over to bash - for the most part, you should avoid it in favour of [[.

e.g.

# With [ - quotes are needed

$ foo='one two'; bar='one two'; [ $foo = $bar ] && echo "they're equal"
-bash: [: too many arguments

$ foo='one two'; bar='one two'; [ "$foo" = "$bar" ] && echo "they're equal" 
they're equal

# versus [[ - quotes not needed

$ foo='one two'; bar='one two'; [[ $foo = $bar ]] && echo "they're equal"
they're equal