[SOLVED] command to display a part of the file

linuxandtsm · 03-22-2012, 12:08 PM

Hi all,

Is there any way that i can display a portion of the file ?
For example, i can find a word in the file using "grep" but i want to display all lines in a file after the word that i just grep'ed for ?

sorry if my explanation is confusing.

Thanks in advance!

suicidaleggroll · 03-22-2012, 12:27 PM

The -A option in grep will print N lines after the match. It's a bit of a hack, but

Code:

grep -A 999999999 word file.txt

will print up to a billion lines after the matching line until it hits the EOF.

David the H. · 03-22-2012, 12:27 PM

The exact answer depends somewhat on the exact nature of your data. If you gave us an example of the text we could give more detailed help.

grep has a -A (after) option for printing a number of lines after the match, but it has to be given a fixed value. There are also -B and -C ("before" and "context" ) options to print lines before and on both sides of the match.

More generally, sed can be given an address range to match a line, and print from there to the end of the file.

Code:

sed -n '/pattern/,$ p'

Here are a few useful sed references.
http://www.grymoire.com/Unix/Sed.html
http://sed.sourceforge.net/grabbag/
http://sed.sourceforge.net/sedfaq.html
http://sed.sourceforge.net/sed1line.txt

ac_kumar · 03-22-2012, 12:42 PM

grep -A 5 "Pattern" file
grep tutorial:-
http://www.devdaily.com/unix/edu/examples/grep.shtml

linuxandtsm · 03-22-2012, 02:18 PM

Wow!.....thanks everyone