[SOLVED] Selective grep

linuxromeo · 07-01-2010, 12:44 AM

Hi

I have to grep a certain entry in a file. But the thing is I need to grep only the first and third occurence of that pattern.

Eg: I have to grep "aaaa=" from the following file

aaaa=1
bbbb=2
dddd=2
fffaaaa=7
aaaa=3
ddaaaa=2
eeee=1

Now I only know "aaaa=" to search for "aaaa=1 and aaaa=3"

Thanks in advance for your help..

colucix · 07-01-2010, 12:59 AM

It is not clear if the problem is to match the first and the third occurrences of the pattern in general or if 1st and 3rd are referred to this specific example (for example, the problem could be to match only the entire word and not the pattern as substring of another word). A real example would help to clarify.

Anyway, in the first case you can try something like:

Code:

grep aaaa file | sed -n '1p;3p'

whereas in the latter a simple

Code:

grep -w aaaa file

should do the trick.

linuxromeo · 07-01-2010, 01:11 AM

Thanks a lot problem solved....

colucix · 07-01-2010, 01:24 AM

Quote:

Originally Posted by linuxromeo

Thanks a lot problem solved....

I am just curious what the problem really was...

linuxromeo · 07-01-2010, 02:21 AM

For the time being...I just need to find the first and third pattern only...But I was just curious, if the given file is modified in future and the third pattern which I needed may become a 4th occurence or more...

eg:
aaaa=1
bbbb=2
dddd=2
fffaaaa=7
aaaa=6
wwaaaa=7
aaaa=3
ddaaaa=2
eeee=1

In this example...aaaa=3 has become the 5th occurence of the pattern "aaaa=".
So all I needed was to write a shell script generically for the same..

grail · 07-01-2010, 02:48 AM

You can get rid of grep and just use all sed too:

Code:

sed -n '/aaaa/{1p;3p}' file