Bash script using sed with variables in a for loop?

Question

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

You need double quotes for variable substitution in bash I think - sed -i "s/${oldString}/${newString}/g" $i; — arunkumar
– arunkumar, Commented Aug 11, 2011 at 22:50

Community · Accepted Answer · 2017-05-23 12:04:04Z

34

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

edited May 23, 2017 at 12:04

CommunityBot

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answered Aug 11, 2011 at 23:03

Karoly Horvath

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Conspicuous Compiler · Accepted Answer · 2011-08-11 22:53:47Z

8

Use double quotes so the environmental variables are expanded by the shell before it calls sed:

  sed -i "s/${oldString}/${newString}/g" $i;

Be wary: If either oldString or newString contain slashes or other regexp special characters, they will be interpreted as their special meaning, not as literal strings.

answered Aug 11, 2011 at 22:53

Conspicuous Compiler

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