【问题标题】:Android how to get the street name from an address returned by GeocoderAndroid如何从Geocoder返回的地址中获取街道名称
【发布时间】:2012-09-14 05:47:22
【问题描述】:
我以相反的方式使用Geocoder 从给定的纬度和经度获取地址。
你知道如何从Address获得只有街道名称吗?
Geocoder geocoder = new Geocoder(AutoFinderMapActivity.this);
try {
List<Address> addressList = geocoder.getFromLocation(latitude, longitude, 1);
if (addressList != null && addressList.size() > 0) {
// Help here to get only the street name
String adress = addressList.get(0).get...;
}
} catch (IOException e) {
e.printStackTrace();
}
提前致谢,
【问题讨论】:
标签:
android
reverse-geocoding
【解决方案1】:
我正在寻找同样的东西。我不同意标记为正确的答案。我可能是错的,但似乎“Thoroughfare”(多么古朴的英文术语!)是提供街道名称的字段。比如:
获取地址:
List<Address> addresses = null;
addresses = geocoder.getFromLocation(latitude, longitude,1);
if(addresses != null && addresses.size() > 0 ){
Address address = addresses.get(0);
// Thoroughfare seems to be the street name without numbers
String street = address.getThoroughfare();
}
【解决方案2】:
而您可能从中获取门牌号的函数是getSubThoroughfare().
【解决方案3】:
这是我的代码示例,显示地址、城市等。希望对你有所帮助。
try {
List<Address> addresses;
Geocoder geocoder= new Geocoder(MyactivityName.this);
addresses = geocoder.getFromLocation(Marker.getPosition().latitude,Marker.getPosition().longitude,1);
if(addresses != null && addresses.size() > 0 ){
Address address = addresses.get(0);
String province = addresses.get(0).getAdminArea();
Marker.setSnippet(address.getThoroughfare()+", "+province);
}
} catch (IOException e) {
e.printStackTrace();
}
Marker.showInfoWindow();
//get current Street name
String address = addresses.get(0).getAddressLine(0);
//get current province/City
String province = addresses.get(0).getAdminArea();
//get country
String country = addresses.get(0).getCountryName();
//get postal code
String postalCode = addresses.get(0).getPostalCode();
//get place Name
String knownName = addresses.get(0).getFeatureName(); // Only if available else return NULL
System.out.println("Street: " + address + "\n" + "City/Province: " + province + "\nCountry: " + country
+ "\nPostal CODE: " + postalCode + "\n" + "Place Name: " + knownName);
如果您查看更多信息或示例,请查看Link