Question

I have a bash script, replace.sh with the following contents:

ack-grep -a -l -i --print0 --text "$1" | xargs -0 -n 1 sed -i -e 's/$1/$2/g'

When I try and run it as, eg:

replace.sh something somethingnew

The prompt returns without errors but no changes have been made to any files.
If I manually type:

ack-grep -a -l -i --print0 --text "something" | xargs -0 -n 1 sed -i -e 's/something/somethingelse/g'

The files get changed as expected.

Ths $1 syntax seems to work for other scripts I’ve written. I’m guessing I’m missing something to do with escaping the args or something?

Thanks!

Ludo.

Answer 1

Variable substitutions aren’t done in single quotes, try:

ack-grep -a -l -i --print0 --text "$1" | xargs -0 -n 1 sed -i -e "s/$1/$2/g"

See the bash man page section on QUOTING.