Why doesn't command substitution work inside `find`?

Question

In bash, if try to run the function basename within the -exec option of find, it doesn't seem to do anything. As an example:

find . -name "*pdf" -exec  echo $(basename {}) \;

yields the filename including the directory, i.e., the same as the result of

find . -name "*pdf" -exec  echo {} \;

Why does this happen?

Answer 1

The $(basename {}) snippet is parsed by your shell before find is executed. The result is {}. That's why the two commands are the same.

To make it work as you wanted you may spawn another shell that will handle $() on its own. Quoting with '' prevents the outer shell from processing $():

find -name "*pdf" -exec sh -c 'echo $(basename "$1")' sh {} \;

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^{Note: the first version of this answer used sh -c 'echo $(basename {})'. This is wrong, see this other answer.}