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display portion of a file

Terminal - display portion of a file
cat -n FILE | grep -C3 "^[[:blank:]]\{1,5\}NUMBER[[:blank:]]"
2009-05-17 18:19:55
User: lv4tech
Functions: cat grep
-1
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display portion of a file

This is useful for displaying a portion of a FILE that contains an error at line NUMBER

Add to favourites | Report as malicious

Know a better way?

If you can do better, submit your command here.

What others think

does not work if you are looking for a line >99 as the line will start with only 3 spaces. Grep options -A and -U can be summarized with -C (context). [[:blank:]] = \s

cat -n FILE | grep -C3 "^\s*NUMBER"
Comment by tomdesinto 209 weeks and 4 days ago

Hmm... My man page says -U means treat files as binary.

-B gives context lines before the match.

Alternatively just replace -A3 -B3 with -C3

Next, you whitespace will only match line numbers from 10-99.

You need ^[[:blank:]]\{1,5\} to match the first 9 lines, and lines 100+.

Finally, 1 will match lots of lines, so you need some more after the number to prevent it.

cat -n FILE | grep -C3 "^[[:blank:]]\{1,5\}NUMBER[[:blank:]]"
Comment by flatcap 209 weeks and 4 days ago

thanks faltcap I've changed it :)

Comment by lv4tech 209 weeks and 4 days ago

What about something like:

sed -n '5,10 p' FILE
Comment by dfego 209 weeks and 3 days ago

Your point of view

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