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Home/ Questions/Q 6097003
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T12:59:34+00:00 2026-05-23T12:59:34+00:00

I have a code like that var=before echo $someString | sed ‘$someRegex’ | while

  • 0

I have a code like that

var="before"  
echo "$someString" | sed '$someRegex' | while read line 
do
    if [ $condition ]; then
        var="after"
        echo "$var" #first echo
    fi 
done 
echo "$var" #second echo

Here first echo print “after”, but second is “before”. How can I make second echo print “after”. I think it is because of pipe buy I don’t know how figure out.

Thanks for any solutions…

answer edit:

I corrected it and it works fine. Thanks eugene for your useful answer

var="before"  
while read line 
do
    if [ $condition ]; then
        var="after"
        echo "$var" #first echo
    fi 
done < <(echo "$someString" | sed '$someRegex')
echo "$var" #second echo
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T12:59:35+00:00Added an answer on May 23, 2026 at 12:59 pm

    The reason for this behaviour is that a while loop runs in a subshell when it’s part of a pipeline. For the while loop above, a new subshell with its own copy of the variable var is created.

    See this article for possible workarounds: I set variables in a loop that’s in a pipeline. Why do they disappear after the loop terminates? Or, why can’t I pipe data to read?.

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