获得企业名称或者标志性建筑从Android的纬度/经度 [英] Getting business names or landmarks from lat/long in Android

问题描述

好了，我一直在寻找在这相当长的一段时间。

我有一个经/纬度从我的新发现的位置。地理codeR的getFromLocation从纬度/长返回一定数量的地址，这是所有罚款和花花公子。然后，我把它放到它填充一个微调适配器...也可以完美运行。

我现在需要的是什么，虽然是让企业的名称，而不是地址。我想preFER使用纬度长，并得到所有的企业在一定距离内，因为不是所有的getFromLocation返回的地址是完整的。有些人只是社区或国家。

此环节做得很好，虽然我还没有实现它。在我看来，这似乎是一个很大的处理能力，什么应该从一个像谷歌大假发pretty标准。

http://www.smnirven.com/?p=39

先谢谢了。

 公共无效getFeature（）{
    双纬度= currentLocation.getLatitude（）;
    双LON = currentLocation.getLongitude（）;
    地理codeR GC =新的地缘codeR（getBaseContext（），Locale.getDefault（））;    尝试{
        清单＆LT;地址＆gt; tmpfeatures = gc.getFromLocation（纬度，经度，15）;
        迭代器ITR = tmpfeatures.iterator（）;
        而（itr.hasNext（））{
            featuresAdapter.add（（（地址）itr.next（））getFeatureName（）的toString（））;
        }
    }赶上（IOException异常五）{
        e.printStackTrace（）;
    }
}
 

解决方案

谷歌Places API的。注册开发人员preVIEW。

HTTP：//$c$c.google.com/apis /地图/文档/地/

Ok, so I have been searching for quite some time on this.

I have a lat/long from my new found location. geocoder's getFromLocation returns a certain number of addresses from the lat/long and that is all fine and dandy. I then put it into an adapter which populates a spinner... also works perfect.

What I need now though is to get business names instead of addresses. I would prefer to use the lat long and get all businesses within a certain distance because not all the addresses returned from getFromLocation are complete. Some are just neighborhoods or countries.

This link does a good job although I have not implemented it yet. In my mind, it seems like a lot of processing power for what should be pretty standard from a big-wig like Google.

http://www.smnirven.com/?p=39

Thanks in advance.

public void getFeature() {
    double lat = currentLocation.getLatitude();
    double lon = currentLocation.getLongitude();
    Geocoder gc = new Geocoder(getBaseContext(), Locale.getDefault());

    try {
        List<Address> tmpfeatures = gc.getFromLocation(lat, lon, 15);
        Iterator itr = tmpfeatures.iterator();
        while (itr.hasNext()) {
            featuresAdapter.add(((Address) itr.next()).getFeatureName().toString());
        }
    } catch (IOException e) {
        e.printStackTrace();
    }
}

解决方案

Google Places API. Sign up for the developer preview.

http://code.google.com/apis/maps/documentation/places/

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