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Old 11-25-2004, 09:18 PM   #1
provkitir
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Registered: Jul 2004
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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
 
Old 11-25-2004, 10:02 PM   #2
mjrich
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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 10:05 PM.
 
Old 11-25-2004, 10:08 PM   #3
whansard
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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.
 
Old 11-25-2004, 10:29 PM   #4
mjrich
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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.
 
Old 11-25-2004, 10:32 PM   #5
provkitir
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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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