Hey everyone! I am determining the best method to do what the subject of this thread says. I only have pieces to the puzzle right now. Namely this:

grep -rl "expression" . | xargs open

(I should mention that the intention is to grep through many files containing the "expression" and return the files themselves for subsequent editing.)

and this:

... | for line in source; do sed 's/expression/replacement/g' > tmp; done

Except that the issue is how to open each file, substitute, and save to the same respective files, not just save to one big tmp file. :P This is eluding me. I realize:

mv tmp > original_file

can be done for each case, but this seems to require a level of scripting knowledge I currently lack. Thanks for any help/suggestions/advice on this!

perl -i.bak -pe 's/this/that/g' <list of file names>

> does not append, > overwrites. So, for each file, overwrite the temp file, then overwrite the original.

If you have GNU sed, you can sed -i to edit in-place instead of cat-ing the new file atop the old.

I don't like the look of this, editing your originals is dangerous, so as a first step BACK UP YOUR FILES. One program bug could wipe out the data of interest here.

# Step 1:  Back up your files.  Editing your originals is dangerous!
grep -rl "expression" . | xargs tar -zcf backup-$(date +%Y-%m-%d-%H-%M-%S).tar.gz

# Step 2:  Find the files, read them in order, substitute.
grep -rl "expression" | while read FILENAME
do
        sed 's/..../' "$FILENAME" > /tmp/$$
        cat /tmp/$$ > "$FILENAME"
done

rm -f /tmp/$$