Re: Finding an Address

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Serge,
	That is precisely what I want!  Any ideas on how to accomplish that?

Thanks!
Floyd


On Feb 28, 2013, at 2:52 PM, Serge Fonville <serge.fonville@xxxxxxxxx> wrote:

> HI,
> 
> It seems like you want something according to the following
> 
> you know your start long/lat
> you can determine the long/lat arround it
> for every of those you determine the route.
> if you follow that route you know the house you find
> otherwise you can use an increasing circle and if it finds an address on the location, you may be able to determine which of the points in the circles (which increase in size) is closest.
> 
> Does that match what you want?
> If not, could you further elaborate what you want exactly?
> 
> Kind regards/met vriendelijke groet,
> 
> Serge Fonville
> 
> http://www.sergefonville.nl
> 
> Convince Microsoft!
> They need to add TRUNCATE PARTITION in SQL Server
> https://connect.microsoft.com/SQLServer/feedback/details/417926/truncate-partition-of-partitioned-table
> 
> 
> 2013/2/28 Floyd Resler <fresler@xxxxxxxxxxxxx>
> 
> 
> 
> On Feb 28, 2013, at 1:04 PM, kenrbnsn@xxxxxxxxx wrote:
> 
> > On 28.02.2013 12:36, Floyd Resler wrote:
> >> I have a project where my client would like to find the nearest
> >> street address from where he current is.  Getting the longitude and
> >> latitude is easy enough but I'm having a hard time finding out how to
> >> get the nearest house.  I have found a lot of solutions for addresses
> >> maintained in a database but these addresses won't be in a database.
> >> I thought about just querying Google for each longitude and latitude
> >> within in a small circle but my math skills are nowhere near good
> >> enough to accomplish that.  Anyone have any ideas?
> >>
> >> Thanks!
> >> Floyd
> >
> >
> > Have you tried Google Maps reverse geocoding? https://developers.google.com/maps/documentation/geocoding/#ReverseGeocoding
> >
> > Ken
> >
> That's what I'm doing but I need to find the closest say five houses to the current latitude and longitude coordinates.
> 
> Thanks!
> Floyd
> 
> 


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