Replace random string starting and ending using shell command

Question

Appreciate any help to find out the solution for below scenario.

Input:

:50Kfoobar:53B:53B:53Bbaz:53B:53B;:50Kvoocar:53B:53B:53Bbaz:53B:53B

Output:

:50KCREDIT:53B:53B:53Bbaz:53B:53B;:50KCREDIT:53B:53B:53Bbaz:53B:53B

Tried using below command

sed 's/:50K.*:53B/:50KCREDIT:53B/g' <<<':50Kfoobar:53B:53B:53Bbaz:53B:53B;:50Kvoocar:53B:53B:53Bbaz:53B:53B'

producing below result
:50KCREDIT:53B

Explanation of @anubhava's suggestion: your `.*` is *greedy*, and will match **everything** up to the last 538. The `[^:]*:` makes the match stop at each colon. — tink, Apr 21 '21 at 07:02
FYI please!! attached dupe could be taken as a reference but doesn't look like an exact dupe. — RavinderSingh13, Apr 22 '21 at 08:29

score 3 · Accepted Answer · answered Apr 21 '21 at 07:02

3

With your shown samples, please try following.

awk '{gsub(/:50K[^:]*:53B/,":50KCREDIT:53B")} 1' Input_file

Explanation: using regex :50K[^:]*:53B which will match :50K till next occurrence of :(non-greedy match) then matching : followed by 53B, then replacing this matched part with :50KCREDIT:53B globally, finally printing the current line.

answered Apr 21 '21 at 07:02

RavinderSingh13

130,504
14
57
93

1

Thanks Ravinder for your time and help. – Sen Apr 21 '21 at 08:14

Replace random string starting and ending using shell command

1 Answers1