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awk: need to extract a line before a pattern

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# 1  
Old 08-02-2006
awk: need to extract a line before a pattern
Hello ,

I need your help to extract a line in a big file , and this line is always 11 lines
before a specific pattern . Do you know a way via Awk ?

Thanks in advance

npn35
# 2  
Old 08-02-2006
mate try this,

awk 'BEGIN {}^J{ if( NR == 11 ) print $0 };^JEND{} ' datafile

Cheers,
# 3  
Old 08-02-2006
Quote:
Originally Posted by nilesrex
mate try this,

awk 'BEGIN {}^J{ if( NR == 11 ) print $0 };^JEND{} ' datafile

Cheers,

In fact , I don't want to print the line 11 . I need to search for a specific pattern
then print the line NR-11 .
I hope my explanation is better .

Thanks.
BR
# 4  
Old 08-02-2006
Try this...

/pattern/ && NR == 11 {
print $0;
}
# 5  
Old 08-02-2006
Quote:
Originally Posted by suman_jakkula
Try this...

/pattern/ && NR == 11 {
print $0;
}
Thanks but the pattern is not located on line 11 . I want to print the line
which is 11 lines before the pattern .

Regards
npn35
# 6  
Old 08-02-2006
Bug Hi
hey man u r not so clearabout what u want?cleat itmate.then it would be easier to give u thesolution.
# 7  
Old 08-02-2006
Quote:
Originally Posted by xinusys
hey man u r not so clearabout what u want?cleat itmate.then it would be easier to give u thesolution.
Quote:
Originally Posted by npn35
I want to print the line which is 11 lines before the pattern .
sounds clear to me.
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