I am trying to search for a shell script if it is running and then assigning the count to a variable. Let the script be ScriptXYZ.sh. I wrote:

proc_count=$(/bin/ps -ef | grep ScriptXYZ.sh | grep -v grep | wc -l)

The problem that I am facing is that even when there is no instance of the ScriptXYZ.sh running I get a proc_count = 2. I tried running the command directly in the terminal and it yielded zero. But this variable always fetches value 2. I tried to put the quotes around it.

proc_count="$(/bin/ps -ef | grep ScriptXYZ.sh | grep -v grep | wc -l)"

But that also didn't change the value. Can anyone please explain why is this happening?

Hello,

The first thing to do is to run the command and see what the output is. When I run the following on my system I get the 0 you are looking for:

[rod@hpsrvr ~]$ /bin/ps -ef | grep ScriptXYZ.sh | grep -v grep | wc -l
0

There is no reason that this should not work.
There must be something else in you program that it does not like.
I used your example but checked for httpd instead of a script and th ran fine.

[rod@hpsrvr ~]$ cat test.sh
#!/bin/bash
myvar=$( /bin/ps -ef | grep httpd | grep -v grep | wc -l)
echo $myvar
[rod@hpsrvr ~]$ ./test.sh
12

Post your code and maybe we will see something.

hi,

pgrep -c XYZ.sh

that's it! XD

Just to add some information that I am runnig it like:

>bash ScriptXYZ.sh [parameters]

and pgrep doesn't show the script as process.

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