Sed substitution does not work with shell variables

Question

I use sed to replace strings in all files.

I have this, but it doens't work:

#!/bin/bash
OLD="oldtext"
NEW="newtext"

grep -rli '$OLD' * | xargs -i@ sed -i 's/$OLD/$NEW/g' @

But this

grep -rli 'oldtext' * | xargs -i@ sed -i 's/oldtext/newtext/g' @

works! How can I fix my command?

Answer 1

As pointed out in randomir's comment and the question he links to, you have to use double quotes so variables are expanded:

old="oldtext"
new="newtext"
grep -rZli "$old" | xargs -0 sed -i "s/$old/$new/gi"

I changed a few other things as well:

• Lowercase variable names to avoid clashes with names reserved for environment variables
• grep -Z option to separate filenames with a NUL byte to deal with special characters such as newlines in filenames
• Skipped * after the sed pattern, as it defaults to the working directory anyway
• -0 option for xargs to expect NUL byte separated filenames
• Skipped the (deprecated) -i option as the filenames go to the end of the command anyway
• i flag for the substitution to match the case insensitive matching in the grep command

Some of these require GNU grep, xargs and sed.

If you don't look at a huge number of files to justify grepping them first, you can just run a single sed process over them all, using a generalized **/* glob:

shopt -s globstar
sed -i "s/$old/$new/gi" **/*

This will complain about sed trying to run on directories, but it doesn't hurt; if all your files have a common extension like .txt, you could use **/*.txt to avoid that.

If, on the other hand, you have tons of files and you want to accelerate the task, you could use the -P option of xargs to parallelize:

grep -rZli "$old" | xargs -0 -P 4 sed -i "s/$old/$new/gi"

for four parallel processes.