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Shell Programming and Scripting

Getting output a line after specific line

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# 1  
Old 01-14-2010
Getting output a line after specific line
Hi

I am a rookie and I would like to know how to make a script that prints as a output lines between two lines let say first variable is a :


Mon Aug 31 09:00:00 2009 CET

and the last line is

Tue Sep 08 09:00:00 2009 CET

also

lets say third line from

Mon Aug 31 09:00:00 2009 CET




thanks in advance I would realy apreciate the help
# 2  
Old 01-14-2010
Quote:
Originally Posted by DinoFly
Hi

I am a rookie and I would like to know how to make a script that prints as a output lines between two lines let say first variable is a :


Mon Aug 31 09:00:00 2009 CET

and the last line is

Tue Sep 08 09:00:00 2009 CET
try:
Code:
sed -n '/Mon Aug 31 09:00:00 2009 CET/,/Tue Sep 08 09:00:00 2009 CET/p' file

or

Code:
awk '/Mon Aug 31 09:00:00 2009 CET/,/Tue Sep 08 09:00:00 2009 CET/' file

# 3  
Old 01-15-2010
Hi and thank you for your fast response.

This works perfectly for the first part but my problem is not yet solved couse I need the solution for printing a specific line after a matched line...,

I want to thank you for quick reply with the solution.
# 4  
Old 01-15-2010
Quote:
Originally Posted by DinoFly
Hi and thank you for your fast response.

This works perfectly for the first part but my problem is not yet solved couse I need the solution for printing a specific line after a matched line...,

I want to thank you for quick reply with the solution.
Try:
Code:
awk '/pattern/{next;print;exit}' file

or:

Code:
sed -n '/pattern/{n;p;}' file

Last edited by Franklin52; 01-15-2010 at 03:42 AM.. Reason: adding sed solution
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