Re: NSSplitView divider position
Re: NSSplitView divider position
- Subject: Re: NSSplitView divider position
- From: Luther Baker <email@hidden>
- Date: Sat, 08 Nov 2014 12:37:11 -0600
Thanks Ken.
On Sat, Nov 8, 2014 at 9:02 AM, Ken Thomases <email@hidden> wrote:
> On Nov 8, 2014, at 8:19 AM, Luther Baker <email@hidden> wrote:
>
> > Is there a direct way to determine the positions of the dividers in
> > NSSplitView.
>
> No.
>
> > If not, are most people calculating this by adding the widths of all the
> > "visible" child/container views to the left of the divider index? (I've
> > noticed the the widths of "collapsed" views in the split view still have
> > the original size).
>
> Well, you can do that, although you also need to account for the
> dividerThickness for any intervening dividers. You could also query the
> NSMaxY() or NSMaxX() of the preceding visible view. (Actually, be careful
> of the user interface direction. Vertical split views that use auto layout
> respect the user interface direction, so the views may be laid out right to
> left.)
>
> > Also, it seems that the NSRect returned from this method is interpreted
> > more like an iOS rect ... that is, Y and the height seem to be top down
> and
> > not bottom up. Is that because the NSSplitView's internal coordinate
> system
> > is managed as such since the docs say ...
> >
> > "The rectangle should be expressed in the coordinate system defined by
> > *splitView*."
>
> You should not assume either way. You should either a) consult the
> -isFlipped method on the split view, or b) construct a rect in the
> coordinate system of one of your own views and then convert to the split
> view's coordinate system using [theSubview convertRect:theRect
> toView:theSplitView] or [theSplitView convertRect:theRect
> fromView:theSubView].
>
> Regards,
> Ken
>
>
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