I'm searching for all files that contain one of a set of strings. If they do, I change their access rights.
find . -type f -exec grep -q '#!/bin/bash\|#!/usr/bin/grep\|#!/usr/bin/awk' {} \; -exec chmod 700 {} ;\
I'm curious how the find works when it finds a file containing that string grep -q gives me 0 so another exec executes but should not. That second exec finds all files and change the rights to all files. Why does it work and change only those files containing that string?
I am not sure what your intention is (you didn't make that clear), but if it's to chmod to 700 all the files that match the pattern, then, except for your typo (;\ instead of \;), your command seems to work as intended.
However:
when it finds a file containing that string
grep -qgives me0so another exec executes but should not.
Yes, it should do. 0 means success means true, so find continues to evaluate its expression, and the second -exec runs. 1 (or any value other than 0) would mean failure would mean false, so find would short-circuit its evaluation (there is an implicit logical AND between juxtaposed expressions) so the second -exec does not run.
Why does it work
Check the manpage for GNU find:
expr1 expr2
Two expressions in a row are taken to be joined with an implied "and"; expr2 is not evaluated if expr1 is false.
... -exec grep -E -q '#! *(/usr)?(/local)?/bin/([bd]ash|[zk]sh|t?csh|awk|sed|perl)' ...(note: I removed the pattern that matches/usr/bin/grepbecause that makes no sense, grep is not a scripting language. Added dash, zsh, ksh, tcsh/csh, sed and perl instead).#!makes more sense than trying to guess al possible interpreters.#!is both at the start of the line, AND is the first line of the file.