Get a specific line from an output using SED in linux

Question

I'm doing:

$> df -kh | sed -n '/\/data\/WAS\/7.0\/wasap03/p'    
/dev/xxxxx/xxxxx          5.0G  490M  4.2G  11% /data/WAS/7.0/wasap03    
/dev/xxxxx/xxxxxxx  3.5G   72M  3.3G  3% /data/WAS/7.0/wasap03/xxxx/xxxx    
/dev/xxxxxx/xxxxxxxxxx  3.5G   72M  3.3G 3% /data/WAS/7.0/wasap03/xxxx/xxxx/xxxx    

$> df -kh | sed -n '/\/data\/WAS\/7.0\/wasap03/p' | awk {'print $5, $6'}    
11% /data/WAS/7.0/wasap03    
3% /data/WAS/7.0/wasap03/xxxxx/xxxxx    
3% /data/WAS/7.0/wasap03/xxxxx/xxxx/xxxxx

My question is I only need the whole line containing only /data/WAS/7.0/wasap03 i.e the output should be only -> 11% /data/WAS/7.0/wasap03, nothing more than this. Please do not say use head -1, I need this only through sed.

@fedorqui I'm sure that this kind of question has been asked before but I don't think that the duplicate you have proposed is correct. This question isn't about getting the nth line, it's about matching an exact pattern. — Tom Fenech, Feb 13 '15 at 12:10
@TomFenech yep I guess you are right. Since the OP stated the `head -1`, I could imagine it was clear to him what line he was looking for. I am reopening it, then I won't be able to cast any close / duplicate vote, so please go ahead if you find an apropiate one. — fedorqui, Feb 13 '15 at 12:12
Also, if you are looking for the `df` info of a specific filesystem, why don't you just do `df -kh /data/WAS/7.0/wasap03`? And then use `awk` to print the desired fields. — fedorqui, Feb 13 '15 at 12:15
possible duplicate of [Sed match exact](http://stackoverflow.com/questions/16196185/sed-match-exact) — Tom Fenech, Feb 13 '15 at 12:16

William Pursell · Accepted Answer · 2015-02-13T12:34:19.990

1

If you want to match the end of line:

df -kh | sed -n '\@/data/WAS/7.0/wasap03$@p'

With awk (so you can avoid piping to awk to get fields 5 and 6:

df -kh | awk 'match($0,"/data/WAS/7.0/wasap03$"){print $5, $6}'

edited Feb 13 '15 at 12:34

answered Feb 13 '15 at 12:26

William Pursell

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very good analyze, i miss it and you make the solution simple and efficient – NeronLeVelu Feb 13 '15 at 12:47
Will, can u pls explain a bit whats the @ and the $ is doing actually in the first command? – Subhadeep Ghosh Feb 13 '15 at 14:58
Subhadeep Ghosh: the `$` matches the end of line. `@` is just being used as an alternate separator so that the `/`'s don't need to be escaped in the pattern. – William Pursell Feb 13 '15 at 17:49

score 0 · Answer 2 · answered Feb 13 '15 at 12:02

0

You could use the below sed command to print only the first match,

$ df -kh | sed -n '0,/\/data\/WAS\/7.0\/wasap03/p'
/dev/xxxxx/xxxxx          5.0G  490M  4.2G  11% /data/WAS/7.0/wasap03

Through awk,

$ df -kh | awk '/\/data\/WAS\/7.0\/wasap03/{print;exit}' 
/dev/xxxxx/xxxxx          5.0G  490M  4.2G  11% /data/WAS/7.0/wasap03
$ df -kh | awk '/\/data\/WAS\/7.0\/wasap03/{print $5,$6;exit}'
11% /data/WAS/7.0/wasap03

answered Feb 13 '15 at 12:02

Avinash Raj

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Beacuse you're new to SO, i suggest you to read [this](http://stackoverflow.com/help/accepted-answer) – Avinash Raj Feb 13 '15 at 12:14

score 0 · Answer 3 · answered Feb 13 '15 at 12:24

df -kh | sed -n '/\/data\/WAS\/7.0\/wasap03/ {s/.*[[:space:]]\([0-9]*%\)/\1/p;q;}'
11% /data/WAS/7.0/wasap03

i suggest to use awk instead because i assume % is alone in the line where awk work with field (same kind of probleme with space in path but you can assume not due to your filter)

Get a specific line from an output using SED in linux

3 Answers3