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Home/ Questions/Q 6560749
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T13:30:05+00:00 2026-05-25T13:30:05+00:00

dpkg –list |grep linux-image |grep ii | while read line do arr=(${line}) let i=i+1

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dpkg --list |grep linux-image |grep "ii  " | while read line
do
  arr=(${line})
  let i=i+1
  _constr+="${arr[2]} "
done
echo $i
echo ${_constr}

The echo statements outside of the loop do not display the expected variables.

How should I make the contents of the variable propagate outside the loop?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T13:30:06+00:00Added an answer on May 25, 2026 at 1:30 pm

    The problem is the pipe, not the loop. Try it this way

    let i=0
    arr=()
    _constr=
    
    while read -r line ; do
        arr=("${line}")
        let i=i+1
        _constr+="${arr[2]} "
    done < <(dpkg --list | grep linux-image | grep 'ii  ')
    
    echo "$i"
    echo "${_constr}"
    

    Pipes are executed in a subshell, as noted by Blagovest in his comment. Using process substitution instead (this is the < <(commands) syntax) keeps everything in the same process, so changes to global variables are possible.

    Incidentally, your pipeline could be improved as well

    dpkg --list | grep '^ii.*linux-image'
    

    One less invocation of grep to worry about.

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