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Pattern output

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# 8  
Old 11-06-2014
Hello znesotomayor,

Could you please try following and let me know if this helps.

Code:
awk '/\/Sun\/Cellular\/version1\/[0-9]*/ {print $0;getline;print $0}' Input_file

Output will be as follows.
Code:
GET /Sun/Cellular/version1/12
HTTP/1.0 100 Internal Error
GET /Sun/Cellular/version1/13
HTTP/1.0 100 Internal Error

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
znesotomayor
# 9  
Old 11-06-2014
How about - if your grep allows for the -B option -
Code:
grep -B1 Error file1
GET /Sun/Cellular/version1/12
HTTP/1.0 100 Internal Error
--
GET /Sun/Cellular/version1/13
HTTP/1.0 100 Internal Error

---------- Post updated at 10:05 ---------- Previous update was at 10:03 ----------

or:
Code:
awk '/Error/ {print T; print} {T=$0}' file1
GET /Sun/Cellular/version1/12
HTTP/1.0 100 Internal Error
GET /Sun/Cellular/version1/13
HTTP/1.0 100 Internal Error

This User Gave Thanks to RudiC For This Post:
znesotomayor
# 10  
Old 11-06-2014
Wow thank you very much Sir RavinderSingh13 and RudiC it works now.. wew at last... thank you very much all for your support. i think i need to learn more from you guys xD thank u thank u...

BR,
# 11  
Old 11-06-2014
Please be aware that - due to your somewhat unspecific specification - the two proposals are NOT equivalent. RavinderSingh's looks for "...version1/something" and prints two lines from there, no matter if an error exists. Which is fine and fulfills the spec. Mine looks for "Error" and prints this along with the line before it, no matter what the version is.

So - in future specs you should be way clearer about what you really need!
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