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Old 08-11-2005, 12:51 AM   #1
hoffmanyew
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Registered: Aug 2004
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Question bash scripts


I wanted to create a script where it will grep the service, output it into a file and kill the services as listed and print it as a log into other file.

How can i capture only the right name services as other might have the same name. For example,

ps -ef | grep vi

root 2138 1 0 09:11:41 ? 0:00 vi something1
root 2709 1072 0 09:13:16 ? 0:01 bvia
root 1511 1466 0 09:10:40 ? 0:04 asbvi

In this case, i just wanted to capture the 'vi' = vi and nothing else...... Can anyone provide a guidance or hints or anything deal with this script?

thanks
 
Old 08-11-2005, 12:55 AM   #2
hk_linux
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try grepping for whole words
Code:
grep -w
HTH
 
Old 08-11-2005, 01:17 AM   #3
hoffmanyew
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thanks for reply. But in this case, it will still capture the other words contain into it i.e

root 2709 1072 0 09:13:16 ? 0:01 bvia
root 1511 1466 0 09:10:40 ? 0:04 asbvi
root 1511 1466 0 09:10:40 ? 0:04 grep -w vi

The idea is that i want to capture service listing with a spawn child which contain 'vi' and not like ivit, vite or anything.
As you can see, it will output the grep and other stuffs. Any better idea how i can capture exact service name 'vi'?

Any suggestion are fully welcome.

thank you
 
Old 08-11-2005, 01:27 AM   #4
hk_linux
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Sorry. I did not fully get what you are saying.

Is it something like,
you want to find a service which is running and send a kill signal to the process.?

In that case you can use "pidof" to find the pid of the process and kill it

Code:
/sbin/pidof PROCESS_NAME
 
  


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