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Home/ Questions/Q 6246863
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T12:46:02+00:00 2026-05-24T12:46:02+00:00

I’m trying to write a bash script that takes a few variables and then

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I’m trying to write a bash script that takes a few variables and then does a find/replace with a given file search using grep to get the list of files that have the string. I think the issue I’m having is having the variables be seen in sed I’m not sure what else it might be.

if [ "$searchFiles" != "" -a "$oldString" != "" -a "$newString" != "" ]; then
   echo -en "Searching for '$searchFiles' and replacing '$oldString' with '$newString'.\n"
   for i in `grep $oldString $searchFiles |cut -d: -f1|uniq`; do
      sed -i 's/${oldString}/${newString}/g' $i;
   done
   echo -en "Done.\n"
else
   usage
fi
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-24T12:46:03+00:00Added an answer on May 24, 2026 at 12:46 pm

    use double quotes so the shell can substitute variables.

    for i in `grep -l $oldString $searchFiles`; do
      sed -i "s/${oldString}/${newString}/g" $i;
    done
    

    if your search or replace string contains special characters you need to escape them: Escape a string for a sed replace pattern

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