Counting number of processes using shell script

Question

I am trying to write a shell script that looks if a particular program is running and if it is not, it restarts the service. Also it checks if the interface is in promiscuous mode. If it is not, it puts it in that mode.

To check if prog xyz is running, I can do the following on commandline

ps -ef | grep -v grep | grep xyz | wc -l

If it returns 0, progrma is not running, else it is running

Similarly to check if the interface is in promisc mode I can do the following on the command line

ip link show eth0 | grep -i promisc | wc -l

Again if return is 1, interface is in promisc mode.

The problem comes when I try to bundle it all in a shell script.

#!/bin/bash
SERVICE="daemonlogger"
x=$(ps -ef|grep -v grep|grep $SERVICE|wc -l)
if [ "$x" -eq 1 ]; then
 run=0
 else
 run=1
fi
IF_NAME="eth0"
y=$(ip link show $IF_NAME|grep -i promisc|wc -l)
if [ "$y" -eq 1 ]; then
 :
 else
 ifconfig $IF_NAME promisc up
fi
if [ "$run" -eq 1 ]; then
 service $SERVICE restart > /dev/NULL 2>&1
fi
echo $run

The output of commands if run from the command line is

[root@dheerajpc Desktop]# ps -ef | grep -v grep | grep daemonlogger | wc -l
0
[root@dheerajpc Desktop]# ip link show eth0 | grep -i promisc | wc -l
0

Here's the output of running it in debug mode

[root@dheerajpc Desktop]# bash -x check_daemonlogger 
+ SERVICE=daemonlogger
++ ps -ef
++ wc -l
++ grep daemonlogger
++ grep -v grep
+ x=2
+ '[' 2 -eq 1 ']'
+ run=1
+ IF_NAME=eth0
++ grep -i promisc
++ ip link show eth0
++ wc -l
+ y=0
+ '[' 0 -eq 1 ']'
+ ifconfig eth0 promisc up
+ '[' 1 -eq 1 ']'
+ service daemonlogger restart
+ echo 1
1

As can be seen the output of first command is not what is expected, while the output of second command is correct.

What am I doing wrong here?

Sounds like you're assuming that only one process can have that in the command line. — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Apr 16, 2012 at 5:58
That's not the point of the question. What I am asking is why does command line return 0 while shell script returns 2? Is there a syntactical problem? — RedBaron
– RedBaron, Commented Apr 16, 2012 at 6:00
Because your "grep daemonlogger" is also a process and will show up in the list of processes. — Noufal Ibrahim
– Noufal Ibrahim, Commented Apr 16, 2012 at 6:00

Ignacio Vazquez-Abrams · Accepted Answer · 2012-04-16 06:00:57Z

2

Hint: What's your script called?

answered Apr 16, 2012 at 6:00

Ignacio Vazquez-Abrams

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Comments

Noufal Ibrahim · Accepted Answer · 2012-04-16 06:02:18Z

1

Not exactly an answer but a bunch of comments.

You should consider using a pidfile rather than relying on ps, grep, wc etc. to track the process. Write the pid to a file and use that to check if the process is running. You've already hit errors with your current approach.
Why do you want to check the status of the interface? Can't you just switch it into promiscuous mode anyway? I haven't played with this but this is the first thing that occurred to me.

answered Apr 16, 2012 at 6:02

Noufal Ibrahim

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