find by name or by grep

Question

find . -perm -u+x,g+x,o+x -type f -exec grep -i -l -H -E '#!/bin/bash|#!/bin/sh' {} \; 2>/dev/null 1>/$user_scripts

I am tasked with finding any script that could exist for N users in 4 folders. I will be using ssh and executing the command above to generate a list of files for later use. Some scripts with the file extension .sh and .bash which do not contain #!/bin/bash|#!/bin/sh are not found.

I need to improve upon this but I am struggling to find something that works in the Solaris server which I have no admin rights for (No special software / libraries). If possible I would prefer not to need to execute another script that would needlessly appear in each list for each user.

Basically I need some kind of "If(find or grep)" in the same line. Is this possible?

Answer 1

You can do this by OR-ing different parts of your find command. Consider splitting your selection criteria into three parts: necessary requirements for all files (permissions, etc), and then EITHER named files OR those with an appropriate header.

This can be expressed with find like this (split onto multiple lines here for readability only):

find . -perm -u+x,g+x,o+x -type f \(
    -name '*.sh' -o
    -name '*.bash' -o
    -exec grep -il -E '^#!/bin/bash|^#!/bin/sh' {} \;
\) 2>/dev/null >/$user_scripts