quick grep question

provkitir · 11-25-2004, 08:18 PM

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

mjrich · 11-25-2004, 09:02 PM

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

whansard · 11-25-2004, 09:08 PM

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.

mjrich · 11-25-2004, 09:29 PM

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.

provkitir · 11-25-2004, 09:32 PM

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!