Grep --only-match without new line

Question

I want to use grep command to extract some strings from my data file.

data file format

a=1,b=2,c=3,
a=4,b=5,c=6,

I want the out format

a=1,c=3,
a=4,c=6,

I tried the below command

grep -Po 'a=.*?,|c=.*?,' data

But the output format is incorrect. Please teach me how to do it.

I suggest to use `cut` or `awk` or `sed`. – Cyrus Jan 06 '18 at 15:22 — Cyrus, Jan 06 '18 at 15:22
awk -F"," '{ print $1","$3"," }' – olivierg Jan 06 '18 at 20:29 — olivierg, Jan 06 '18 at 20:29

score 0 · Answer 1 · answered Jan 06 '18 at 16:58

0

One possibility with sed would be as follows.

sed -n 's/\(a=.\?,\)\(.*\)\(c=.\?\)/\1\3/p' data

A more generic regex would be as follows.

sed -n 's/\([a-zA-Z]=.\?,\)\([a-zA-Z]=.\?,\)\([a-zA-Z]=.\?,\)/\1\3/p' input

answered Jan 06 '18 at 16:58

Thomas

1 Answers1