Fdzfgy

Honestly I don't even know how to ask this, or what are the proper terms lol

So what I'm trying to do, is to work on variables with variable names. I think I'd best describe it by actually showing it:
So what I had before was this

if [ -z ${file_var} ]; then

        echo "empty argument"

fi

if [ -z ${pw_var} ]; then

        echo "empty argument"

fi

So what I was trying to do was not have two if statements, but have just one and a for loop, something like this:

for i in file_var, pw_var;do

        if [ -z ${${i}} ]; then

                echo "empty argument"

        fi

done

Now this obviously doesn't work, so I'm wondering how is it properly done.

What you're trying to do is indirect expansion, which Bash supports using !:

if [ -z "${!i}" ] ; then

If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new parameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter. This is known as indirect expansion.

In this case, "${!i}" expands to whatever the value of the variable whose name is stored in i, and -z will test that indirectly-accessed value.

Alternatively, a nameref is a special type of variable that has this behaviour automatically. You make a nameref variable using declare -n name, and thereafter

All references, assignments, and attribute modifications to name, except for those using or changing the -n attribute itself, are performed on the variable referenced by name’s value.

So if you add

declare -n i

before your loop, it will mean that just

for i in file_var pw_var

do

    if [ -z "$i" ]

    then

    ...

    fi

done

will work as you wanted. Do note though that assignments to i (as opposed to its being set in a loop like this) will modify the target variable.

Note also that there is no comma between items you're looping over in a Bash for loop like how you've written it in the question.

This code works, if file_var or pw_var are empty it will print that the specific variable is empty. You can tweak it to fit your usage.

file_var=""

pw_var=""



for i in 'file_var' 'pw_var';do

  if [ -z ${!i} ]; then

    echo "$i is empty"

  fi

done