cat all lines from file, which are after "KEYWORD"

Question

I have a file, with a "KEYWORD" on line number n. How can I print all lines starting from line n+1 until the end?

For example, here I would liek to pro=int only lines DDD and EEE

AAA
BBB
CCC
KEYWORD
DDD
EEE

I could see this being really useful for log files with, say, something like a time stamp as the keyword. — Joel Coehoorn, Jun 23 '13 at 17:52

Chris Down · Accepted Answer · 2015-06-24T08:10:07.143

28

You can do this with sed:

sed '1,/^KEYWORD$/d'

This will delete (omit) all lines from the beginning of the stream until "KEYWORD", inclusive.

edited Jun 24 '15 at 08:10

answered Jun 23 '13 at 08:29

Chris Down

5

For me this didn't work and I had to use sed '1,/KEYWORD/d', instead. The context was a little different, but using 0, would print the contents of the entire file, no matter what KEYWORD was. – juniper- Mar 16 '15 at 08:08

suspectus · Answer 2 · 2013-06-23T16:56:50.737

3

Use sed, printing from the KEYWORD match to the end of the file, then using tail to remove the KEYWORD line.

sed -n '/KEYWORD/,$p' file | tail -1

edited Jun 23 '13 at 16:56

answered Jun 23 '13 at 08:31

suspectus

score 3 · Answer 3 · answered Jun 23 '13 at 09:52

An alternative might be grep in combination with the -A flag, e.g.

grep -A10000 KEYWORD file

Where 10000 is just a big number to denoted the amount of lines until the end of the file, which for your practical daily use should be enough.

Otherwise you could use the amount of lines in the file as parameter like this

grep -A$(wc -l file | cut -d' ' -f1) KEYWORD file

But that is most likely overkill (and not easier to remember than the given sed alternative)

You'd need the | tail -1 trick of @suspectus to omit the first line. — Bernhard, Jun 23 '13 at 09:55

3 Answers3