I have one script that only writes data to stdout. I need to run it for multiple files and generate a different output file for each input file and I was wondering how to use find -exec for that. So I basically tried several variants of this (I replaced the script by cat just for testability purposes):
find * -type f -exec cat "{}" > "{}.stdout" ;
but could not make it work since all the data was being written to a file literally named{}.stdout.
Eventually, I could make it work with :
find * -type f -exec sh -c "cat {} > {}.stdout" ;
But while this latest form works well with cat, my script requires environment variables loaded through several initialization scripts, thus I end up with:
find * -type f -exec sh -c "initscript1; initscript2; ...; myscript {} > {}.stdout" ;
Which seems a waste because I have everything already initialized in my current shell.
Is there a better way of doing this with find? Other one-liners are welcome.
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Answer
A simple solution would be to put a wrapper around your script:
#!/bin/sh myscript "$1" > "$1.stdout"
Call it myscript2 and invoke it with find:
find . -type f -exec myscript2 {} ;
Note that although most implementations of find allow you to do what you have done, technically the behavior of find is unspecified if you use {} more than once in the argument list of -exec.