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Cut third field of every third line

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# 1  
Old 05-24-2011
Cut third field of every third line
Hello,

I have got a log file and would need to write a script to cut the every first and second fields of every third line.

Job Name : dummytextd_v1
Status : KILLED
TIMEDOUT 2011-05-01 05:33
Job Name : dummyttx_v1
Status : KILLED
TIMEDOUT 2011-05-03 02:33

Job Name : dummy_v1
Status : KILLED
TIMEDOUT 2011-05-03 03:33

Job Name : dummy_v1
Status : KILLED
TIMEDOUT 2011-05-03 10:33

So in the above example, I just need to cut the TIMEOUT and Time (2011 *) . Can someone could help me with this? :-)
# 2  
Old 05-24-2011
What is the desired output for this data?
# 3  
Old 05-24-2011
I guess you are looking for line TIMEOUT, which is not exactly every third line.

Try anyone of this.

Code:
cgi@tonga> (/home/cgi) $ awk '/TIMEDOUT/{printf "%s %s\n", $1, $2 }' test.txt
TIMEDOUT 2011-05-01
TIMEDOUT 2011-05-03
TIMEDOUT 2011-05-03
TIMEDOUT 2011-05-03
cgi@tonga> (/home/cgi) $ awk '{ if(NR==int((NR/4))*4+3){printf "%s %s\n",$1, $2} }' test.txt
TIMEDOUT 2011-05-01
TIMEDOUT 2011-05-03
TIMEDOUT 2011-05-03
TIMEDOUT 2011-05-03
cgi@tonga> (/home/cgi) $

# 4  
Old 05-24-2011
it seems u want all the lines that begin with TIMEDOUT
if that is the case then u can use this command
Code:
grep -i "TIMEDOUT" file.txt

the contents of file.txt wud be something like this :
TIMEDOUT 2011-05-01 05:33
TIMEDOUT 2011-05-03 02:33
TIMEDOUT 2011-05-03 03:33
and so on .

hope this helps Smilie..
Last edited by vbe; 05-24-2011 at 06:22 AM.. Reason: code tags ...
# 5  
Old 05-24-2011
oh thanks to both Pratham and Kumaran
Code:
awk '{ if(NR==int((NR/4))*4+3){printf "%s %s\n",$1, $2} }' test.txt

worked perfectly :-)

Other methods could be used, but in this case its just not the "TIMEDOUT" on every third(fourth) line, SUCCEEDED and KILLED too should be cut. :-) thanks again for your help.
Last edited by vbe; 05-24-2011 at 06:22 AM.. Reason: please use code tags!
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