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

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
>
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Are you looking for:
:echo expand("%:r")

:help filename-modifiers

Best,
Marcin

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