Monitoring shell script

dsevern · 01-05-2011, 03:35 PM

I'm having to write a script to make up for a command line products shortfalls.

The command will kick off multiple background processes, the number of processes is based on the number of filenames in a given file; workspace.lst.

Each process will create a logfile with the same name as the input filename in a directory; LOG_DIR.

I don't know how long it will take each process to complete, but for this exercise I can assume no more than 1 hour.

I need to grep each log filename in workspaces.lst in LOG_DIR for the existance of a "Success" string. If it exists, processing for that filename is done and check the others. If it doesn't I want to wait 5 minutes and check again. If after 60 minutes all are complete, exit. Or if after 60 minutes some logfiles do not contain the success string, assume failure.

I can't even seem to pseudo-code this, much less bash script it. If anyone is bored and has some time, your help would be greatly appreciated.

Snark1994 · 01-05-2011, 05:36 PM

I think this should work...

Code:

#!/bin/bash

x=11
LOG_DIR="/here/be/logs"
cd $LOG_DIR
while [ $x -ge 0 ] 
do
	finished=1
	for f in *
	do
		grep Success "$f" >/dev/null
		if [ "$?" -gt 0 ]; then
			finished=0
		fi
	done
	
	if [ "$finished" -eq 1 ]; then
		echo "Success"
		exit 0
	fi

	sleep 300 
	let x-=1
done

kill -9 `ps aux | grep the_process | egrep -v grep | awk '{print $2}'`
echo "Failure"
exit 1

EDIT: What does it say about my life that the first thing I thought when I woke up this morning was "I missed a line in the code I wrote last night?". The script has been amended to kill all of the processes before reporting failure.