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Editorial Team
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Asked: 2026-06-15T14:06:33+00:00

I am trying to parse a filename from a modified apache web access log

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I am trying to parse a filename from a modified apache web access log entry that is tab delimited:

/common/common/img/pictos/klArrowRight.gif    /common/common/img/pictos/klArrowRight.gif   03/Dec/2012:00:00:00    127.0.0.1   03/Dec/2012:00:00:00    us   404

I would like it to come out like this:

klArrowRight.gif    /common/common/img/pictos/klArrowRight.gif   03/Dec/2012:00:00:00    127.0.0.1   03/Dec/2012:00:00:00    us   404

I have tried something like this in :

‘s:.*/::’

However, it is too greedy, and it eats the rest of my line. I have been looking through posts, but so far no luck. Any hints?

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1 Answer

  1. Editorial Team

    the input/output in your question is not well formatted. do you need this?

    awk '{gsub(/\/.*\//,"",$1); print}' file

    test

    kent$  echo "/common/common/img/pictos/klArrowRight.gif /common/common/img/pictos/klArrowRight.gif 03/Dec/2012:00:00:00 127.0.0.1 03/Dec/2012:00:00:00 us 404"|awk '{gsub(/\/.*\//,"",$1); print}'

    output:

    klArrowRight.gif /common/common/img/pictos/klArrowRight.gif 03/Dec/2012:00:00:00 127.0.0.1 03/Dec/2012:00:00:00 us 404
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