Bash Question
shell script gives output in 2 lines I need only 1
I have the following commands saved in a .sh file
prog=$1
ps axf | grep $prog | grep -v grep | awk '{print "kill -9 " $1}'
I get the following output when I execute it
kill -9 3184
kill -9 20359
But I just need the first line of it as that is the only valid pid. How can I remove the 2nd line from the output.
Answer
There are a few issues with what you want to do:
- You're building a chain of 4 commands for something relatively simple
- You're going to get as a result only the first line of a list of processes containing
$prog(excluding thegrep $progwhich you filtered out); how can you be sure that's the process you want?
The correct command to use is
pkill $prog`
as suggested in the comments, which probably will do what you want.
Just for information, and to answer your question, you can pipe an output to head -n 1 to return only the first line:
<list of commands> | head -n 1
However, in your case this would add a fifth command to the chain, so I recommend you don't do it this way.