Regex to replace a string

Question

In VI I need to replace a pattern that looks like this:

seq- followed by exactly 24 chars (can be 0-9, a-z, A-Z only) followed by / followed by some number of (0-9, a-z, A-Z). So an example string would be:

seq-BzFk78qrJBkbnv1KkxppsfG4/92191_A_B

I want to replace the above string with a fixed string:

seq-ID/NUM_ID

How do I do this in Vi or Vim? I tried a few things like

  :%s/seq-[a-zA-Z0-9]{24}/[a-zA-Z0-9]{+}/seq-ID/NUM_ID

But it doesnt' work..

Cross Post: vi.stackexchange.com/q/9690/46

FDinoff
– FDinoff

2016-09-27 02:16:24 +00:00
Commented Sep 27, 2016 at 2:16 — FDinoff
– FDinoff, Commented Sep 27, 2016 at 2:16

redneb · Accepted Answer · 2016-09-27 00:28:23Z

1

You have to be careful with how you use quantifier and you must escape the /. Try this

:%s/^seq-[a-zA-Z0-9]\{24\}\/.*$/seq-ID\/NUM_ID/

answered Sep 27, 2016 at 0:28

redneb

24.3k7 gold badges46 silver badges55 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Regex to replace a string

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related