Vim Regex Replace

Question

I think this is a pretty simple solution but I have been trying to find an answer for about an hour and can't seem to figure it out.

I'm trying to do a find/replace in Vim that would replace a series of numbers that begin with 103 and are followed by four digits to 123 followed by the same four digits, as such:

1033303 -> 1233303
1033213 -> 1233213

The closest I have come is this:

%s/103\d\{4}/123\0/g

However this results in the entire number being matched by the \0, as such:

1033303 -> 1231033303

Thank you very much for any help

beware chaining regex greediness in vim is weird vi.stackexchange.com/questions/196/… — ThorSummoner
– ThorSummoner, Commented Dec 30, 2019 at 3:22

Conner · Accepted Answer · 2012-08-07 16:20:06Z

37

You're very close

%s/103\(\d\{4}\)/123\1/g

The pattern between \( and \) is a sub-match that can be accessed by \1, \2 etc in the order of appearance. See :help \( for more information.

edited Aug 7, 2012 at 16:20

answered Aug 7, 2012 at 16:16

Conner

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Comments

Peter Rincker · Accepted Answer · 2012-08-07 16:44:50Z

10

Now w/o the capture group!

:%s/103\ze\d\{4}/123/g

\ze will set the end of the match there. This allows for you match the whole pattern but only do the substitution on a portion of it.

:h /\ze
:h /\zs

answered Aug 7, 2012 at 16:44

Peter Rincker

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3 Comments

SergioAraujo Over a year ago

Would .* after \ze do the trick once everuthing after it os ignored?

Peter Rincker Over a year ago

Yes, it would prevent more matches on the same line

SergioAraujo Over a year ago

This worked here: :%s,103\ze.*,123,g changing delimiter (fast to type) and matching all subsequent chars with .* which will be ignored on substitution.

Andrew Clark · Accepted Answer · 2012-08-07 16:17:50Z

4

Use a capturing group:

%s/103\(\d\{4}\)/123\1/g

answered Aug 7, 2012 at 16:17

Andrew Clark

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Comments

srini.venigalla · Accepted Answer · 2012-08-07 16:18:47Z

2

Try this:

%s#103\(\d\{4}\)#123\1#g

answered Aug 7, 2012 at 16:18

srini.venigalla

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