Kill a list of processes


 
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# 1  
Old 10-04-2014
Kill a list of processes

I am trying to kill a list of processes. I have found these two ways to list a group of process id's on a single line. How would I go about killing all of these processes all on one line?

Code:
$ ps aux | grep 6243 | grep "a.out" | awk '{printf "%s ",$2}'

Code:
ps aux | grep 6243 | grep "a.out" | awk '{print $2}' ORS=' '

Code:
7543 7553 7558 7561 7569 7574


Last edited by cokedude; 10-05-2014 at 01:28 AM..
# 2  
Old 10-04-2014
Huh? You don't have a list of PIDs; you have one PID listed 5 times.
Since all you seem to be interested in is PID 6243, why not just use:
Code:
kill 6243

???
# 3  
Old 10-05-2014
Quote:
Originally Posted by Don Cragun
Huh? You don't have a list of PIDs; you have one PID listed 5 times.
Since all you seem to be interested in is PID 6243, why not just use:
Code:
kill 6243

???
I fixed my mistake. I was doing a lot of copying and pasting. $1 was supposed to be $2. 6243 was the uid I was using to filter the user I wanted.

Mistake
Code:
$ ps aux | grep 6243 | grep "a.out" | awk '{printf "%s ",$1}'

Corrected
Code:
$ ps aux | grep 6243 | grep "a.out" | awk '{printf "%s ",$2}'

# 4  
Old 10-05-2014
Quote:
Originally Posted by cokedude
I fixed my mistake. I was doing a lot of copying and pasting. $1 was supposed to be $2. 6243 was the uid I was using to filter the user I wanted.

Mistake
Code:
$ ps aux | grep 6243 | grep "a.out" | awk '{printf "%s ",$1}'

Corrected
Code:
$ ps aux | grep 6243 | grep "a.out" | awk '{printf "%s ",$2}'

So, if:
Code:
$ ps aux | grep 6243 | grep "a.out" | awk '{printf "%s ",$2}'

gives you the list of PIDs you want to kill, it is a pretty small step from there to:
Code:
$ kill $(ps aux | grep 6243 | grep "a.out" | awk '{printf "%s ",$2}')

isn't it?
# 5  
Old 10-05-2014
Quote:
Originally Posted by cokedude
[...] 6243 was the uid I was using to filter the user I wanted.
As a way of information `ps' has some more options that might allow you to tailor more custom output that just `ps aux'.

For example if you want to kill any processes containing "a.out" for the user 6243:
Code:
kill $(ps -U 6243 | awk '$5 ~ /a\.out/ {print $1}')

As a "rule of thumb" anything that grep can do, awk can also achieve. So if you use awk, you most likely do not have to bother with another filter like grep.

There's also a very useful unix utility named `lsof' that can be handy when dealing with processes. If your system has it installed it is worthwhile to get familiar with it.
If you want to kill only the processes number of the program top that user 6243 has initiated.

Code:
kill $(lsof -t -u 6243 -a -c top)

-t: will output only the pid
-u: user id or name
-a: ANDing operator
-c: command

Last edited by Aia; 10-05-2014 at 02:57 AM.. Reason: Adding meaning of flags
# 6  
Old 10-05-2014
Quote:
Originally Posted by Don Cragun
So, if:
Code:
$ ps aux | grep 6243 | grep "a.out" | awk '{printf "%s ",$2}'

gives you the list of PIDs you want to kill, it is a pretty small step from there to:
Code:
$ kill $(ps aux | grep 6243 | grep "a.out" | awk '{printf "%s ",$2}')

isn't it?
Yep thats it Smilie. Thank you. I wasn't familiar with that $() trick. Where can I read more about things like that?

---------- Post updated at 02:30 AM ---------- Previous update was at 02:08 AM ----------

Quote:
Originally Posted by Aia
As a way of information `ps' has some more options that might allow you to tailor more custom output that just `ps aux'.

For example if you want to kill any processes containing "a.out" for the user 6243:
Code:
kill $(ps -U 6243 | awk '$5 ~ /a\.out/ {print $1}')

As a "rule of thumb" anything that grep can do, awk can also achieve. So if you use awk, you most likely do not have to bother with another filter like grep.

There's also a very useful unix utility named `lsof' that can be handy when dealing with processes. If your system has it installed it is worthwhile to get familiar with it.
If you want to kill only the processes number of the program top that user 6243 has initiated.

Code:
kill $(lsof -t -u 6243 -a -c top)

-t: will output only the pid
-u: user id or name
-a: ANDing operator
-c: command
This worked when I changed it.

Code:
kill $(ps -U 6243 | awk '$4 ~ /a\.out/ {print $1}')

ps -U only has 4 columns.

Can you please explain this part? I don't understand the purpose of a "~" here and why you had to the a.out in "//" . Would bash try to expand the "." if you didn't "\" it?

Code:
~ /a\.out/

The system I am using is old and unfortunately doesn't have lsof.
# 7  
Old 10-05-2014
Quote:
Originally Posted by cokedude
Yep thats it Smilie. Thank you. I wasn't familiar with that $() trick. Where can I read more about things like that?
man bash (or whatever your shell be) is your friend:

Quote:
Command Substitution
Command substitution allows the output of a command to replace the command name. There are two forms:

$(command)
or
`command`
Quote:
Originally Posted by cokedude
. . .
Can you please explain this part? I don't understand the purpose of a "~" here and why you had to the a.out in "//" . Would bash try to expand the "." if you didn't "\" it?
. . .
man awk is your friend:
Quote:
5. Expressions and operators
The expression syntax is similar to C. . . .

New expressions are composed with the following operators in order of increasing precedence.

assignment = += -= *= /= %= ^=
.
.
.
matching ~ !~
.
.
.
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