Multiple find -exec commands in one bash script doesn't work?

Question

I have a bash script that needs to be run by cron. It works when the script only contains 1 command line, but fails when it's more than 1 line.

#!/bin/sh
find /path/to/file1 -name 'abc_*' -type f -mtime +7 -exec rm {} \;
find /path/to/file2 -name 'def*.gz' -type f -mtime +7 -exec rm {} \;

I received find: missing argument to `-exec' error message. I need to keep only the last 7 days of several different files in several different directories.

Why did I get that error message when all the commands have already seem to be true?

Answer 1

@user1576748

Is there anything that would prevent you from doing this inside one line?

example:

find /path/to/file1 /path/to/file2 -name 'abc*' -o -name 'def*.gz' -type f -mtime +7 -exec rm {} \\;

The above works for me.