passing variable to this particular sed command

Question

I am first searching for a key word and once that key word is found in a file from that particular line i am supposed delete till end of file.

#! /bin/csh -f
set sa = `grep -n -m 1 "^Pattern" file`
set s = `echo "$sa" | cut -d':' -f1`   
set m = `sed '$s,$d' file | tee see > /dev/null`

so first line gives me the matching line with line number, second line i am getting the line number and third line i am trying to delete from line $s say 20 till last but it is not working. I have tried all combinations but it does not take the variable $s. Please help.

You know that variables are only expanded inside double quotes, not inside single quotes, don't you? This is Shell Scripting 101. — Barmar, Jun 13 '14 at 07:12
Better question: why are you writing a script in csh? http://www.faqs.org/faqs/unix-faq/shell/csh-whynot/ — Barmar, Jun 13 '14 at 07:21

score 2 · Answer 1 · answered Jun 13 '14 at 07:17

2

But you can do it much more easier with a single line of sed:

sed -n '/SEARCHPATTERN/q;p

-n tells to not print the lines
/SEARCHPATTERN/q exits on search pattern
;p otherwise print the lines

answered Jun 13 '14 at 07:17

Zsolt Botykai

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score 1 · Accepted Answer · answered Jun 13 '14 at 07:23

1

You need to take $s out of the quotes so it will be expanded.

set m = `sed $s',$d' file | tee see > /dev/null`

answered Jun 13 '14 at 07:23

Barmar

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Sorry for tagging bash. It worked. Actually this is one of the several part of script which i am implementing in csh. – user2508758 Jun 13 '14 at 07:29

passing variable to this particular sed command

2 Answers2