Is it possible to get “find -exec” expand backticks for each file found?

Question

Here's a command that works fine:

echo Blah: `stat -c %a .`

Is it possible to get find -exec to execute this same thing for every file found, with {} in place of .?

Answer 1

The best way to do that is to use positional parameters. And $() is preferred for command substitution ove backticks because it's more readable (isn't confused with single quotes) and can be easily nested without having to do a lot of escaping.

find . -exec bash -c 'echo Blah: $(stat -c %a "$@")' _ {} \;

The underscore is a placeholder for $0.

Answer 2

What for?

find ... -printf 'Blah: %m\n'

Or use bash -c if you really have to:

 find . -exec bash -c 'echo Blah: `stat -c %a {}`' \;

Answer 3

The only way to use them (AFAIK) is to place them in a bash script file and use the script in -exec. Same thing with $()

Answer 4

Don't forget xargs ! Your example would invoke a new shell process for each file found.

I would prefer this:

find ... | xargs stat -c "Blah: %a"

find outputs a list of everything found, xargs reads a list of arguments on stdin and executes its parameter with those arguments on the command line, building as long a command line as possible. It works because stat, as most other proper commands/programs do, take any number of parameters. (compare ls, rm and echo for instance)

If you think it's absolutely neccessary to launch a new process for each file, use xargs -n 1 to only pass 1 parameter to each command. That way you can mimick the inefficient methods like this:

find ... | xargs -n 1 stat -c "Blah: %a"

(Try it on a big filesystem on a slow computer and time the differences!)