Delete the lines before the first instance of the keyword

Shell Programming and Scripting
1

I have my data something like this. I want to delete all the lines before the frist instance of the key word 'ravi kumar'

aaa
bbbbbb cccccc
ddddd eeeee
1234 ravi kumar
aaaaaa vvvvvvv
5678 ravi kumar
rrrrrrr mmmmmmm

I want the output as follows.

1234 ravi kumar
aaaaaa vvvvvvv
5678 ravi kumar
rrrrrrr mmmmmmm

Please help me.

2 
awk '/ravi kumar/,/*/ {print $0}'  yourfle
3

How about this?

 
 sed -n '/ravi kumar/,/*/p' file
4

Sed is working perfectly where as there seems to be a syntax issue with awk command above.
Thanks a lot for your help.

5 
sed -n '/ravi kumar/,$p' file

awk '/ravi kumar/,i++' file

Jean-Pierre.

6

Jean-Pierre,

Your awk command ignores the last line.

Another approach with awk:

awk '/ravi kumar/{p=1}p' file

Regards

7

Oups! you're right Franklin.

awk '/ravi kumar/,p' file

Input:

aaa
bbbbbb cccccc
ddddd eeeee
1234 ravi kumar
aaaaaa vvvvvvv
bbbbbbb wwwwwwww
ccccccccc zzzzzzzzzz
5678 ravi kumar
rrrrrrr mmmmmmm END

Output:

1234 ravi kumar
aaaaaa vvvvvvv
bbbbbbb wwwwwwww
ccccccccc zzzzzzzzzz
5678 ravi kumar
rrrrrrr mmmmmmm END

Jean-Pierre.

8

Thanks a lot for your inputs. Similarly, how can i delete the lines after the last occurence of the keyword?

9

Bumping up posts or double posting is not permitted in these forums.

Please read the rules, which you agreed to when you registered, if you have not already done so.

Proceed here:

http://www.unix.com/shell-programming-scripting/115621-delete-lines-after-last-instance-keyword.html\#post302338987

Thank You.

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