Grep empty space and sort

Shell Programming and Scripting
1

Hi Expert,

Kindly request for your expertise in this matter.

I have below output:

12.125.124.173,xx1.common.com
12.125.124.174,xx2.common.com
12.125.124.175,xx3.common.com
12.125.124.176,
12.125.124.177,
12.125.124.178,
12.125.124.179,xx4.common.com
12.125.124.180,xx5.common.com
12.125.124.181,xx6.common.com
12.125.124.182,xx19.common.com
12.125.124.183,xx54.common.com
12.125.124.184,
12.125.124.185,
12.125.124.193,
12.125.124.194,
12.125.124.195,
12.125.124.196,xx33.common.com
12.125.124.197,xx89.common.com
12.125.124.198,xx88.common.com

and how do I script it to just appear with:
by omitting the xxx.common.com

12.125.124.176,
12.125.124.177,
12.125.124.178,
12.125.124.184,
12.125.124.185,
12.125.124.193,
12.125.124.194,
12.125.124.195,

thank you.

2 
grep -v "common" yourinputfile
3

As you mentioned about sort,

grep -v "common" file|sort
4

thanks for reply.

but how do I just grep blank instead of "common"?
I would like to have this output

12.125.124.176,
12.125.124.177,
12.125.124.178,

instead of

12.125.124.173,xx1.common.com
12.125.124.174,xx2.common.com
12.125.124.175,xx3.common.com

Thank you.

5

if the comma is always present at the end of the line you want, use grep ',$'

6

what do you mean balnk, you mean the second field after IP is blank or whole line is blank ?

7

sed -n '/common/!p' <file name>

8

If you want to keep the IP but remove the part containing "common" :

sed 's/,.*/,/' yourfile
9

not quite there yet,

What I wanted is to exclude all lines that include 'common' word.

Thank you,
Paul Albert

---------- Post updated at 10:52 PM ---------- Previous update was at 05:23 PM ----------

Thanks. SOLVED!