vim regex find and replace

Question

I want to use a regex to replace some strings in my file. I search for: %s/^ [a-z]*/ / what I want to do is to replace every [a-z]* that have 2 whitespaces with the sane [a-z] prepended with 4 whitespaces. Is there any "inplace" replacement or how would I reach that with vim?

With best regards

Do you mean replace [][]abcd with [][][][]abcd in which [] stands for space? — amazingjxq
– amazingjxq, Commented Jun 30, 2011 at 10:02

sehe · Accepted Answer · 2011-06-30 09:50:18Z

5

:%s/  \([a-z]*\)/    \1/g

should do the job; beware of running this multiple times, though because the result of the replace will match the input pattern :)

answered Jun 30, 2011 at 9:50

sehe

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1 Comment

Hassek Over a year ago

I am very sad this was down voted, that \1 is priceless! thanks for the awesome tip :)

Prince Goulash · Accepted Answer · 2011-06-30 10:11:22Z

3

I find it more straightforward to use the \ze object to define the end of the match:

:%s/  \ze[a-z]*/    /g

so the [a-z]* is not included in the replace, but just used to match the relevant spaces.

answered Jun 30, 2011 at 10:11

Prince Goulash

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2 Comments

Eugene Yarmash Over a year ago

This will also affect lines, prepended with 4 whitespaces.

sehe Over a year ago

I like that approach. It'll also be more efficient (I'd be surprised if that would ever lead to noticable speed differences, but hey, I'm a programmer).

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