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I use following code to verify if variable is not empty and this works fine

VAR=(var1 var2 var3)

chec_var() {
for item in "${@}"; do
if [ -z "${!item}" ]; then
  returncode=1
  break
else
  returncode=0
fi
done
}

all_vars() {
if chec_var "${VAR[@]}"; then
  if [[ ${returncode} -gt 0 ]]; then
    echo "error"
    exit 1
  else
    echo "OK"
  fi
fi
}

all_vars

The problem stars when I add set -Euo pipefail at the end of the script.
Than when one of variables is not present I get following error:
/script line number: !item unbound variable
but it should display "echo error" instead message from pipefail

Another time when all variables are present than it checks only first variable and displays
"echo OK"
instead of going through the loop and check all variables

Question:
What am I doing wrong here?
How to loop through all variables and once all are check to display OK when -Euo pipefail is enabled?

Modified script with pipefail looks as following:

VAR=(var1 var2 var3)

chec_var() {
for item in "${@}"; do
if [ -z "${!item}" ]; then
  return 1
  break
else
  return 0
fi
done
}

all_vars() {
if chec_var "${VAR[@]}"; then
  if [[ ${returncode} -gt 0 ]]; then
    echo "error"
    exit 1
  else
    echo "OK"
  fi
fi
}

set -Euo pipefail

all_vars
Improve this question
5
  • Are you sure the set line goes at the end of the script? Commented Aug 12, 2021 at 11:35
  • Also, the -E makes the ERR trap inherited by functions. I don't see any trap defined in your script, though. Commented Aug 12, 2021 at 11:41
  • lost {} now this is right example Commented Aug 12, 2021 at 11:41
  • Possible duplicate of How do I check if a variable exists in an 'if' statement? Commented Aug 12, 2021 at 12:44
  • the script itself works fine for me and in the pointed article How do I check if a variable exists in an 'if' statement? there is nothing about set -Euo pipefail Commented Aug 12, 2021 at 17:41

1 Answer 1

Reset to default
1

here's one way to do it in bash:

if declare -p variable >&/dev/null ; then
  echo "variable exists"
else
  echo "variable does not exist"
if

declare -p will return with an exit code of 0 if the variable exists, and 1 if it doesn't. The redirection of stdout and stderr is because we don't want to see any output, we just want to test if the variable exists.

In most cases, though, just testing if the variable is empty (i.e. [ -z "$variable" ] or, invert the test for not-empty, with [ -n "$variable" ]) is good enough.

Improve this answer
4
  • [ -n "$variable" ] or [ -z "$variable" ] would trip set -o nounset aka set -u if the variable is unset as that's still a attempt at dereferencing it. Using the ${variable+set} parameter expansion operator cancels the set -o nounset effect. Commented Aug 12, 2021 at 16:17
  • That's part of the reason why I said "in most cases". set -o nounset isn't "most cases". Commented Aug 12, 2021 at 16:42
  • Yes, but the question is "how do you do it with set -u", so, it's actually "in most cases, but not yours". Commented Aug 13, 2021 at 7:12
  • ya know, that's why it was an aside at the end of my answer, not my actual answer. Commented Aug 13, 2021 at 7:53

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