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Skipping the 1st occurence of a search/only printing the nth?
I need to process a search of an array, locating the nth occurrence of the
search and only printing that particular record.
The following will locate the criteria and print, but only the first
occurence.
nawk '$1 == "RTOUT" {print $0, exit}' ${inputF} > ${outputF}
Any help on skipping the first occurence, locating the nth and only
printing that record?
Any suggestions would be appreciated,
groblela
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groblela
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9/10/2004 2:29:18 PM |
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groblela <lee.grobleski@nospam.wpafb.af.mil> wrote:
> I need to process a search of an array, locating the nth occurrence of the
> search and only printing that particular record.
> The following will locate the criteria and print, but only the first
> occurence.
>
> nawk '$1 == "RTOUT" {print $0, exit}' ${inputF} > ${outputF}
>
> Any help on skipping the first occurence, locating the nth and only
> printing that record?
>
> Any suggestions would be appreciated,
> groblela
Use a counter. If counter reaches 'n', then print.
--
William Park <opengeometry@yahoo.ca>
Open Geometry Consulting, Toronto, Canada
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William
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9/10/2004 2:37:51 PM
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On Fri, 10 Sep 2004 10:29:18 -0400 in comp.lang.awk, "groblela"
<lee.grobleski@nospam.wpafb.af.mil> wrote:
>I need to process a search of an array, locating the nth occurrence of the
>search and only printing that particular record.
>The following will locate the criteria and print, but only the first
>occurence.
>
>nawk '$1 == "RTOUT" {print $0, exit}' ${inputF} > ${outputF}
>
>Any help on skipping the first occurence, locating the nth and only
>printing that record?
nawk -vN=5 '$1 == "RTOUT" { if (++cnt == N) { print $0; exit } }'
--
Thanks. Take care, Brian Inglis Calgary, Alberta, Canada
Brian.Inglis@CSi.com (Brian[dot]Inglis{at}SystematicSW[dot]ab[dot]ca)
fake address use address above to reply
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Brian
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9/10/2004 4:16:46 PM
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2 Replies
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