Redirecting stdout with find -exec and without creating new shell
linux, unix
Solution
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`.
Problem
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.