[mapserver-users] network analysis in mapserver
woodbri at swoodbridge.com
woodbri at swoodbridge.com
Mon Jul 15 07:37:20 PDT 2002
Seamus,
Mapserver doesn't have this functionality as you probably know at
this point. I have code that does what you describe below that could
be adapted to any number of similar application. To be able to run
fast (ie for a web application) I build a directed graph that is then
used to compute the shortest path method and then I extract the
required segments into a shape file that can be rendered by
mapserver.
I have a live demo running on my site at
http://iMapTools.com/demos
click the "Get A Route" link
This demo is using the US National Highway Planning Network
shapefiles for the directed graph, but a directed graph could be
built for other networks from shapefiles or other data sources.
-Stephen Woodbridge
On 12 Jul 2002 at 9:43, Seamus Thomas Carroll wrote:
> Hi,
>
> I have post GIS installed but I am not currently using it. I have a road
> network for British Columbia and some other provinces and two sets of
> point data. I need to be able to find the shortest path along from point
> A (postal code) to point B (hospital data) using an impedance taken from
> the dbf. For example, the length in metres for an arc or the amount of
> time to travel an arc. I need the total impedance and a shape file
> representing the path as a result. Would any of these tools solve this
> problem for me?
>
> Seamus
>
>
> On Fri, 12 Jul 2002, Robert W. Burgholzer wrote:
>
> > Seamus,
> > Are you using postGIS by chance? I am working on a set of tools that
> > might not be network analysis, but we may have some overlap that we
> > could share.
> >
> > I too am seeking independence from esri, and postgres extendability (all
> > functions can be written in C, or several other languages) seems to me
> > to be a potential for high quality, open source GIS tools. Already there
> > are several geo-processing routines out there acting in
> > postgres/postgis. What type of analysis functions are you looking for?
> >
> > Rob
> >
> >
>
>
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