I'm trying to construct a find command to process a bunch of files in a directory using two different executables. Unfortunately, -exec on find doesn't allow to use pipe or even \| because the shell interprets that character first.
Here is specifically what I'm trying to do (which doesn't work because pipe ends the find command):
find /path/to/jpgs -type f -exec jhead -v {} | grep 123 \; -print
As this outputs a list would you not :
find /path/to/jpgs -type f -exec jhead -v {} \; | grep 123
or
find /path/to/jpgs -type f -print -exec jhead -v {} \; | grep 123
Put your grep on the results of the find -exec.
Try this
find /path/to/jpgs -type f -exec sh -c 'jhead -v {} | grep 123 \; -print' \;
Alternatively you could try to embed your exec statement inside a sh script and then do:
find -exec some_script {} \;
With -exec you can only run a single executable with some arguments, not arbitrary shell commands. To circumvent this, you can use sh -c '<shell command>'.
Do note that the use of -exec is quite inefficient. For each file that is found, the command has to be executed again. It would be more efficient if you can avoid this. (For example, by moving the grep outside the -exec or piping the results of find to xargs as suggested by Palmin.)
A slightly different approach would be to use xargs:
find /path/to/jpgs -type f -print0 | xargs -0 jhead -v | grep 123
which I always found a bit easier to understand and to adapt (the -print0 and -0 arguments are necessary to cope with filenames containing blanks)
This might (not tested) be more effective than using -exec because it will pipe the list of files to xargs and xargs makes sure that the jhead commandline does not get too long.