bash script if variable equals

Question

How do i add a variable as the name of the variable i want to checks value. My code is as below, when i run this i get the error:

[: A_variable: integer expression expected

I think this is because when doing an if on a variable the proper syntax should be: if [ $A_variable -eq 1 ]; but i am not able to put the dolla sign in front of my variable because i am putting a different variable in the value of my variable.

Its almost like it should be: if [ $${letter}_variable or some mix of this but i dont know what?

array=("A" "B" "C")

A_variable=1
B_variable=2
for letter in "${array[@]}"
do

if [ ${letter}_variable -eq 1 ]; then
...

You would probably be better off using an associative array, instead of a bunch of similarly named but independent variables. See e.g. my older answer to a similar question and the section on them in BashFAQ. — ilkkachu, Sep 20 '19 at 14:28

telcoM · Accepted Answer · 2019-09-20T13:30:43.490

This is called indirect expansion in bash.

The syntax to get the value of a variable whose name is determined by another variable namevar is:

${!namevar}

Since your variable name is composed of a variable and a constant string, you'll need one intermediate step:

array=("A" "B" "C")

A_variable=1
B_variable=2
for letter in "${array[@]}"
do
    varname="${letter}_variable"
    if [ "${!varname}" -eq 1 ]; then
    ...

Note: if the indirectly-referenced variable does not actually exist, the expansion will result in a null string, which will still cause an error. If this is not desirable, add a second test like this first:

    varname="${letter}_variable"
    if [ "${!varname}" = "" ]; then
        echo "Error: $varname is non-existent or empty"
    elif [ "${!varname}" -eq 1 ]; then
    ...

Also, using double quotes to protect against unexpected characters in the expansion result would be a good idea, if the contents of all the possible matching variables are not initialized to safe values in advance.

This will still throw an error for "C": [: : integer expression expected — markgraf, Sep 20 '19 at 13:09
Well, yes... because in the OP's original setup, array has three elements, A, B and C, and only two of the variables resulting from the expansion have been initialized. I understood the core of the question was about the indirect expansion, and the OP did not indicate what they want to happen if the indirect-referenced variable does not exist. — telcoM, Sep 20 '19 at 13:21

markgraf · Answer 2 · 2019-09-20T13:16:33.077

You are comparing a string ("A_variable") to an integer ("1").

You can get around this issue by using declare -n. From help declare:

-n make NAME a reference to the variable named by its value

We need [[ .. ]] to avoid the error the unset "C_variable" is causing.

#!/bin/bash
array=("A" "B" "C")

A_variable=1
B_variable=2
for letter in "${array[@]}"
do
  declare -n foo="${letter}_variable"
  if [[ "$foo" -eq 1 ]]; then
    echo "Ding"
  fi
done

Bonus: you can declare -i A_variable=1 to make it an integer.

-i to make NAMEs have the `integer' attribute

schrodingerscatcuriosity · Answer 3 · 2019-09-20T12:57:45.607

0

The operator -eq is for math expressions, but you are comparing strings.

To campare strings use =.

edited Sep 20 '19 at 12:57

answered Sep 20 '19 at 12:52

schrodingerscatcuriosity

12,396

I think they're trying to use the value of (say) A_variable, which is indeed a number. – ilkkachu Sep 20 '19 at 14:25

bash script if variable equals

3 Answers3