Can't use a variable out of while and pipe in bash

Question

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

fedorqui · Accepted Answer · 2014-09-16 10:56:38Z

7

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?.

edited Sep 16, 2014 at 10:56

fedorqui

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answered Jul 4, 2011 at 14:12

Eugene Yarmash

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