Get N lines before a pattern

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11-19-2012

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Get N lines before a pattern

I need to print N lines before a pattern ERR-XXXXX .

The grep around and before option is not working for me.

input file

Code:

line 1 some text
line 2 some other text
line 3 ERR-12345
line 4 some more text
line 5 some blah blah blah
line 6 ERR-56789

n=1

Output

Code:

line 2 some other text
line 3 ERR-12345
line 5 some blah blah blah
line 6 ERR-56789

With awk , I am getting the NR , is it possible to print line NR-1 , NR-2 with AWK

Last edited by Scott; 11-19-2012 at 01:44 AM.. Reason: Code tags

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11-19-2012

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Hi

Code:

$ awk '{a[++i]=$0;}/ERR-/{for(j=NR-x;j<=NR;j++)print a[j];}' x=1 file
some other text
ERR-12345
some blah blah blah
ERR-56789

You can specify your value of N against the x.

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Code:

grep -B 1 "ERR"

works for me

Jotne

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11-19-2012

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Rotating version, uses less memory:

Code:

awk '{A[NR%(n+1)]=$0} /ERR-/{for(i=NR-n; i<=NR; i++)print A[i%(n+1)]}' n=1 infile

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