How to use > in an xargs command?
I want to find a bash command that will let me grep every file in a directory and write the output of that grep to a separate file. My guess would have been to do something like this
ls -1 | xargs -I{} "grep ABC '{}' > '{}'.out"
but, as far as I know, xargs doesn't like the double-quotes. If I remove the double-quotes, however, then the command redirects the output of the entire command to a single file called '{}'.out instead of to a series of individual files.
Does anyone know of a way to do this using xargs? I just used this grep scenario as an example to illustrate my problem with xargs so any solutions that don't use xargs aren't as applicable for me.
Answer
Do not make the mistake of doing this:
sh -c "grep ABC {} > {}.out"
This will break under a lot of conditions, including funky filenames and is impossible to quote right. Your {} must always be a single completely separate argument to the command to avoid code injection bugs. What you need to do, is this:
xargs -I{} sh -c 'grep ABC "$1" > "$1.out"' -- {}
Applies to xargs as well as find.
By the way, never use xargs without the -0 option (unless for very rare and controlled one-time interactive use where you aren't worried about destroying your data).
Also don't parse ls. Ever. Use globbing or find instead: http://mywiki.wooledge.org/ParsingLs
Use find for everything that needs recursion and a simple loop with a glob for everything else:
find /foo -exec sh -c 'grep "$1" > "$1.out"' -- {} \;
or non-recursive:
for file in *; do grep "$file" > "$file.out"; done
Notice the proper use of quotes.