How do I surround a line with quotes on bash?

Question

When I run for i in $(find -name '*.ogg'); do echo '$i'; done I don't get one line per file as I expected. Instead, when a file name has a space in it, each word comes out in a separate line, so I can't run commands on the files.

Answer 1

find -name '*.ogg' -print0 | xargs -0 -L1 command

Answer 2

Here is an example of maybe what you want to do:

 #! /bin/sh
 IFS="\n"
 for xxx in `cat /etc/hosts`
 do
    echo $xxx
 done
 exit 0

I.E: Change the IFS (Inter Field Seperator) to \n instead of space,tab,\n

Answer 3

You should pipe find into a while loop instead:

find -name '*.ogg' | while read -r i; do echo "$i"; done

Also, the single quotes you had around $i would prevent the variable from being expanded to its value.

Another way is to use process substitution:

 while read -r i; do echo "$i"; done < <(find -name '*.ogg')

which has the advantage of not creating a subshell out of the while loop, so variables set (and other changes to the environment) inside the loop be available after the loop is finished.

Answer 4

If you want to run the command in parallel and if have GNU Parallel http:// www.gnu.org/software/parallel/ installed you can do this:

find -name '*.ogg' | parallel do_stuff {} --options otherargs

Watch the intro video for GNU Parallel to learn more: http://www.youtube.com/watch?v=OpaiGYxkSuQ