How to perform a sed in-place substitution that only creates backups of files that were changed?

Question

I ran the following to replace a term used in all files in the current working directory:

$ find . -type f -print0 | xargs -0 sed -i'.bup' -e's/Ms. Johnson/Mrs. Melbin/g'

This performed the word substitution but it also created .bup files of files that never had the Ms. Johnson string.

How do I perform the substitution without creating all these unnecessary backups?

Answer 1

You could check the files' contents to make sure that a substitution will take place when sed operates on them:

find . \
    -type f \
    -exec grep -q 'Ms. Johnson' {} \; \
    -print0 |
xargs -0 sed -i'.bup' -e's/Ms. Johnson/Mrs. Melbin/g'

If you want to be really clever about it, you could forego using find at all:

grep -Z -l -r 'Ms. Johnson' |
    xargs -0 sed -i'.bup' -e's/Ms. Johnson/Mrs. Melbin/g'

Answer 2

Find has an -exec operator that executes an arbitrary command. Even better, -exec is a test, so you can chain several -execs together, and if earlier commands fail, later ones won't get executed. The string {} gets replaced with the current filename, and ; marks the end of the command. You should quote both to avoid the shell interfering.

So:

find . -type f \
    -exec grep -q 'Ms. Johnson' '{}' \; \
    -exec sed -i'.bup' -e's/Ms. Johnson/Mrs. Melbin/g' '{}' \;

Answer 3

I looked at the man page and didn't see any way to do it directly through sed, as I'm sure you did before asking. I see several ways to work around this using grep, but I think the easiest is this:

grep -rlZ "Ms. Johnson" . | xargs -0 sed -i'.bup' -e's/Ms. Johnson/Mrs. Melbin/g'

-r recurse
-l print filename only
-Z end names with null

Answer 4

Seems like you need to build a list of files that match your search terms first.

search="test"
for match in $(find -type f -exec grep -l $search "{}" \;)
do sed -i'.bup' -e "/$search/ s/$search/Mrs. Melbin/g" "$match"
done