How to check dynamically that a Bash variable exists? [duplicate]

Question

Please consider the following Bash function.

#!/bin/bash

function assert_var(){
  echo "check if variable with name $1 exists"
  if [ -z $1 ];then
    echo "EXISTS!"
  else
    echo "NOT EXISTS!"
  fi
}


export FOO="123"
assert_var "FOO"
assert_var "BAR"

The expectation is that variable FOO should be detected and absence of BAR as well. For that, I need somehow to pass and use the variable name as argument to the function.

How to do that properly?

Be sure to use -v, though. The more upvoted duplicate is old and has more upvotes for answers that predate -v. — chepner
– chepner, Commented Jul 1, 2020 at 14:16

anubhava · Accepted Answer · 2020-07-01 13:52:53Z

2

This function should do the job for you:

assert_var() {
   if [[ -z ${!1+x} ]]; then
      echo 'NOT EXISTS!'
   else
      echo 'EXISTS!'
   fi
}

Changes are:

Use ${var+x} to check if a variable is unset. ${var+x} is a parameter expansion that evaluates to nothing if var is unset, and substitutes the string x otherwise. More details here.
Use indirect referencing of variable by name i.e. ${!1+x} instead of ${1+x}
Use single quotes when you are using ! in string to avoid history expansion

Testing:

FOO='123'
assert_var "FOO"
EXISTS!
assert_var "BAR"
NOT EXISTS!

edited Jul 1, 2020 at 13:52

answered Jul 1, 2020 at 13:49

anubhava

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2 Comments

0x5453 Over a year ago

Can you explain what ${!1+x} is doing?

anubhava Over a year ago

yes I just added

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