CODEWITHSUNDEEP
regexvimreplace
Glennular
Glennular

Reputation: 18215

VIM Search/Replace spaces between 2 brackets or columns

Given the following line:

[aaaa bbbb cccc dddd] [decimal](18, 0) NULL,

How would you replace the spaces only between the first set of brackets in Vim? What would the /s command look like?

Update:

This is the intend outcome

    [aaaa_bbbb_cccc_dddd] [decimal](18, 0) NULL,

Upvotes: 4

Views: 4112

Answers (3)

thegeek
thegeek

Reputation: 2458

You can always go for interactive find and replace, which can be done with flag c

Upvotes: -1

intuited
intuited

Reputation: 24034

with the normal setting of 'magic' you'd want to do

:s/] \[/][/

or more fancily

:s/]\zs\s\+\ze\[//

or with the magic level explicitly set in the regex

:s/\V]\zs\s\+\ze[//

\zs and \ze limit the substitution to the area that they enclose.

\M turns off magic completely, so every character or character combo not prefixed with \ is taken literally.

'magic' (:help 'magic') is a global option that determines how potentially-special characters in regexes are interpreted.

Upvotes: 1

jamessan
jamessan

Reputation: 42657

The easiest way to do it would be to visually select the contents of the []ed string and perform the replacement just on that selection:

vi]:s/\%V \%V/_/g

This doesn't work very well if you're trying to do things programmatically, though. In that case, you can match the entire []ed string and use a replacement expression to construct the result.

:s/\[[^\]]*\]/\=substitute(submatch(0), ' ', '_', 'g')/g

Upvotes: 7

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