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, Aug 11 '11 at 22:50

score 34 · Accepted Answer · edited May 23 '17 at 12:04

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 '17 at 12:04

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answered Aug 11 '11 at 23:03

Karoly Horvath

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score 8 · Answer 2 · answered Aug 11 '11 at 22:53

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.

Bash script using sed with variables in a for loop?

2 Answers2

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