Find -exec dry run?

Question

Is there a way to see what the result of a find . -exec somecommand {} \; would be with substitutions, without actually running the commands? Like a dry run (or test run or print)?

For example, suppose I have the following file structure:

/a/1.txt
/a/2.txt
/a/b/3.txt

Is there a way to test find . type f -exec rm {} \; from within the a directory such that the output would printed to stdout but not executed such as:

rm 1.txt
rm 2.txt
rm b/3.txt

Update Note: rm is just an example command, I'm interested in the general case

Answer 1

You can run echo rm instead of rm

find . type f -exec echo rm {} \;

Also, find has -delete option to delete files it finds

Answer 2

For rm specifically, you don't need -exec: simply run find . -type f to list, and add -delete to delete the files listed by the previous command (obviously barring any matching files being created/deleted in the meantime).

Also, for commands like rm which take an arbitrary number of arguments you'll want to replace \; with + to run as few commands as possible.

Answer 3

It's a bit of a mouthful, but unlike approaches using echo, the below outputs code you could run in your shell without any changes to have the correct result, even when your filenames contain quotes, spaces, shell metacharacters, etc.

printcmd() { printf '%q ' "$@"; printf '\n'; }

find . -exec bash -c "$(declare -f printcmd); "'printcmd "$@"' _ \
  somecommand {} \;

Note that the string we're prepending to our -exec argument is precisely bash -c "$(declare -f printcmd); "'printcmd "$@"' _ -- the $(declare -f printcmd) expands to the code for the function; after that, we actually call the function with arguments $1 and onward, and put _ as a placeholder for $0.

You can substitute zsh or ksh instead of bash, if you want output escaped for that shell.