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11-25-2004, 08:18 PM
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#1
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Member
Registered: Jul 2004
Location: Mass
Distribution: Freebsd 5.3, Debian sid 2.6.7
Posts: 101
Rep:
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quick grep question
Hi
i have a quick grep question, say there's a big file where there are many instances of "1125" on differnent lines. however, I just want the four letters before 1125, what's the grep command line to show only one of the many instances with only the "YYYY1125" as output?
I read thru man grep and most of which was regarding to the number of lines
thanks
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11-25-2004, 09:02 PM
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#2
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Senior Member
Registered: Dec 2001
Location: New Zealand
Distribution: Debian
Posts: 1,046
Rep:
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Code:
grep -oh [[:space:]]....1125[[:space:]] <datafile>
Or something along those lines may do the trick, depending on how your data file is laid out.
Cheers,
mj
Last edited by mjrich; 11-25-2004 at 09:05 PM.
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11-25-2004, 09:08 PM
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#3
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Senior Member
Registered: Dec 2002
Location: Mosquitoville
Distribution: RH 6.2, Gen2, Knoppix,arch, bodhi, studio, suse, mint
Posts: 3,306
Rep:
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that would require a grep command piped through an awk command. have grep print the line, and have awk chop it down to what you want.
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11-25-2004, 09:29 PM
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#4
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Senior Member
Registered: Dec 2001
Location: New Zealand
Distribution: Debian
Posts: 1,046
Rep:
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The above grep command will work, as long as you happen to have a tab or some other delimiter before the four characters and the 1125, and only one 1125 per line. But that's a big if, I guess !
You could also use head/tail to chop the output, but awk would be more reliable, as per whansard's suggestion.
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11-25-2004, 09:32 PM
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#5
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Member
Registered: Jul 2004
Location: Mass
Distribution: Freebsd 5.3, Debian sid 2.6.7
Posts: 101
Original Poster
Rep:
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cool!
it's working now! thanks a lot!
hehe, comics.com have a new 4 number code before their daily comics everyday and I wanted to use shell to get those comics and now i can
thanks again!
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