how do i parse by specific column?

kmaq7621 · #1 (**permalink**) 09-02-2008

I have a log file with 13 columns. The 12th column contains the status code (example 200, 404, 500, 403, etc.) I want to remove all 200 status lines. so...

1. remove all the lines in which the 12th column has a 200.
2. display only the lines in which the 12th column shows a 500.

Thanks.

danmero · #2 (**permalink**) 09-02-2008

Code:

awk '$12!=200' file
awk '$12==500' file

ShawnMilo · #3 (**permalink**) 09-02-2008

Quote:

Originally Posted by kmaq7621

I have a log file with 13 columns. The 12th column contains the status code (example 200, 404, 500, 403, etc.) I want to remove all 200 status lines. so...

1. remove all the lines in which the 12th column has a 200.
2. display only the lines in which the 12th column shows a 500.

Thanks.

perl -ane 'print unless $F[11] eq "200"' temp.txt
perl -ane 'print if $F[11] eq "500"' temp.txt

Note: The -a flag to Perl is "auto-split," which creates an array named F which you can then address by element.

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