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Can't access variables inside parenthesis in bash

In bash, if I run

(foo=14)

And then try to reference that variable later on in my bash script:

echo "${foo}"

I don't get anything. How can I make bash store this variable the way I need it to?

Specifically, I am using this in an if statement and checking the exit code, something kind of like:

if (bar="$(foo=14;echo "${foo}"|tr '1' 'a' 2>&1)")
then
    echo "Setting "'$bar'" was a success. It is ${bar}"
else
    echo "Setting "'$bar'" failed with a nonzero exit code."
fi
like image 632
leetbacoon Avatar asked Oct 31 '25 14:10

leetbacoon


1 Answers

Commands enclosed in parenthesis e.g. () are executed in a sub-shell. Any assignment in a sub-shell will not exist outside that sub-shell.

foo=14
bar=$(echo $foo | tr '1' 'a' )
if [[ $? -eq 0 ]]
then
    echo "Setting "'$bar'" was a success. It is ${bar}"
else
    echo "Setting "'$bar'" failed with a nonzero exit code."
fi
like image 192
suspectus Avatar answered Nov 02 '25 13:11

suspectus



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