Run a list of bash scripts consecutively

Question

I have a load of bash scripts that backup different directories to different locations. I want each one to run every day. However, I want to make they don't run simultaneously.

I've wrote a script that basically just calls each script in succession and sits in cron.daily, but I want a way for this script to work even if I add and remove backup scripts without having to manually edit it.

So what I need to go is generate a list of the scripts (e.g. "dir -1 /usr/bin/backup*.sh") and then run each script it finds in turn.

Thanks.

Answer 1

#!/bin/sh    
for script in /usr/bin/backup*.sh
do
$script
done

Answer 2

#!/bin/bash
for SCRIPT in /usr/bin/backup*.sh
do
   [ -x "$SCRIPT" ] && [ -f "$SCRIPT" ] && $SCRIPT   
done

Answer 3

If your system has run-parts then that will take care of it for you. You can name your scripts like "10script", "20anotherscript" and they will be run in order in a manner similar to the rc*.d hierarchy (which is run via init or Upstart, however). On some systems it's a script. On mine it's a binary executable.

It is likely that your system is using it to run hourly, daily, etc., cron jobs just by dropping scripts into directories such as /etc/cron.hourly/

Pay particular attention, though, to how you name your scripts. (Don't use dots, for example.) Check the man page specific to your system, since file naming restrictions may vary.