Re: Get filename part only from "% with substitute()

On Sunday, October 14, 2012 4:59:28 PM UTC-7, coot_. wrote:
> On 15:25 Sun 14 Oct , Graham Lawrence wrote:
>
> > :echo substitute(@%, "\..*", "", "")
>
> >
>
> > :echo substitute(@%, "\([^.]\+\)\..*", "\1", "")
>
> > Parameters2.html
>
> >
>
> > >From :reg
>
> > ": echo substitute(@%, "\([^.]\+\)\..*", "\1", "")
>
> > "% Parameters2.html
>
> >
>
> > So, how should I write this substitute command?
>
> >
>
> >
>
> > --
>
> > Graham Lawrence
>
> >
>
> > --
>
> > You received this message from the "vim_use" maillist.
>
> > Do not top-post! Type your reply below the text you are replying to.
>
> > For more information, visit http://www.vim.org/maillist.php
>
>
>
> Are you looking for:
>
> :echo expand("%:r")
>
>
>
> :help filename-modifiers
>
>
>
> Best,
>
> Marcin

Many thanks, that solves the immediate problem perfectly. But I would also like to know why substitute() does not, if possible.

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