[OpenLayers-Users] boundaries of lineString
Tim Schaub
tschaub at openplans.org
Tue Feb 12 10:52:21 EST 2008
Hey-
Rupesh M G wrote:
>
> Thank you for the reply. But I think it is not correct.
>
> For example, if I draw the linestring with 4 points, I'm able to see a
> polygon with 8 vertices.
A linestring geometry with 4 points has 4 points. When you see it
rendered, it looks like a polygon to you, but the geometry is just a
linestring.
It sounds like you're talking about buffering a geometry. When you
buffer a linestring geometry, you get a polygon geometry.
Tim
>
> But if I print the components.length of the linestring, it just says 4.
>
> I tried it with the example in the url
> http://openlayers.org/dev/examples/vector-features.html
>
>
> Regards,
> Rupesh
>
>
>
> *"Pierre GIRAUD" <bluecarto at gmail.com>*
>
> 02/11/2008 03:58 PM
>
>
> To
> "Rupesh M G" <RupeshM at ibsplc.com>
> cc
> users at openlayers.org
> Subject
> Re: [OpenLayers-Users] boundaries of lineString
>
>
>
>
>
>
>
>
> It should be quite easy because geometry like lineString and Polygon
> are composite geometries.
> For example, a OpenLayers.LineString has a 'components' property which
> is an array of OpenLayers.Points.
> That way, you can get all the vertices using a loop like following :
>
> for(var i=0; i <= thelinestring.components.length; i++) {
> // do something with thelinestring.components[i] which is an OL point
> }
>
> Regards,
> Pierre
>
> On Feb 11, 2008 9:02 AM, Rupesh M G <RupeshM at ibsplc.com> wrote:
> >
> > Hi,
> >
> > I'm working on an application about roads. There I need to mark a
> > part of the road, and execute some query.
> >
> > My user will click a set of points to form a sequence of lines passing
> > through the center of the road, because there can be bends in the road.
> >
> > I plan to draw a lineString going through these points. Using a bigger
> > strokeWidth, I can show it like a road (or like a filled polygon).
> >
> > Is there any way to get the vertices of that polygon?
> >
> >
> > Thanks & Regards,
> > Rupesh
> >
> >
> >
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