Re: Substitution words within a partial match range help

On Thursday, April 18, 2013 8:27:36 AM UTC-5, David Fishburn wrote:
> Vim 7.3.758
>
>
> I have the following string:
>
> 'match /\%(if\|elsif\|unless\|given\|when\|default\)\>/ match /\<\>/  nextgroup=perlElseIfError skip0 skip1 skip2'
>
>
>
> I was hoping (using 1 or 2 substitution commands) to find all words less than or equal to 5 characters in length and remove them.
>
>
> But, these words have to be within "match /.../" tags.
>
>
>
> So, when completed, the string above should read:
>
> 'match /\%(\|\|unless\|\|\|default\)\>/ match /\<\>/  nextgroup=perlElseIfError skip0 skip1 skip2'
>
>
>
> Notice the skip0, skip1, skip2 are not touched since they are not within the beginning "match /" and the ending "/" of the match.
>
>
> My best try was the following:
>
> >echo substitute( 'match /\%(if\|elsif\|unless\|given\|when\|default\)\>/ match /\<\>/  nextgroup=perlElseIfError skip0 skip1 skip2', '\<match /.\{-}\zs\<\w\{1,5}\>\ze.\{-}/ ', '', 'g' )
>
>
>
> But it only gets rid of the "if" at the start, and doesn't work on any of the other matches.
>
>
> TIA,
> David

I got it working *better* using \@<= instead of \zs:

\%(\<match /.\{-}\)\@<=\<\w\{1,5}\>\ze.\{-}/

But this also removes the second "match" in the line, so you'll need to refine it some more, using [^/] instead of . to avoid that the ending '/' is allowed to match the beginning pattern:

\%(\<match /[^/]\{-}\)\@<=\<\w\{1,5}\>\ze.\{-}/

Now I think it works as you intended.

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