如何在android中使用纬度和经度获取城市名称
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【中文标题】如何在android中使用纬度和经度获取城市名称【英文标题】：how to get city name using latitude and longitude in android 【发布时间】：2017-10-07 07:23:31 【问题描述】：我正在尝试使用纬度和经度在 TextView 中获取城市。我收到IndexOutOfBoundsException。
AndroidGPSTrackingActivity.java
import android.app.Activity;
import android.location.Address;
import android.location.Geocoder;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;
import android.widget.Toast;
import java.io.IOException;
import java.util.List;
import java.util.Locale;
import static com.example.khaledsb.location.R.id.lng;
import static java.util.Locale.*;
public class AndroidGPSTrackingActivity extends Activity
public static float distFrom(float lat1, float lng1, float lat2, float lng2)
double earthRadius = 6371000; //meters
double dLat = Math.toRadians(lat2-lat1);
double dLng = Math.toRadians(lng2-lng1);
double a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2)) *
Math.sin(dLng/2) * Math.sin(dLng/2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
float dist = (float) (earthRadius * c);
return dist;
Button btnShowLocation;
// GPSTracker class
GPSTracker gps;
@Override
public void onCreate(Bundle savedInstanceState)
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
btnShowLocation = (Button) findViewById(R.id.btnShowLocation);
final TextView lat =(TextView) findViewById(R.id.lat);
final TextView lon = (TextView) findViewById(R.id.longt);
final TextView addr = (TextView) findViewById(R.id.address);
// show location button click event
btnShowLocation.setOnClickListener(new View.OnClickListener()
@Override
public void onClick(View arg0)
// create class object
gps = new GPSTracker(AndroidGPSTrackingActivity.this);
// check if GPS enabled
if(gps.canGetLocation())
double latitude = gps.getLatitude();
double longitude = gps.getLongitude();
double Distance;
Distance = distFrom((float) 36.5925907, 2.9051544f, 36.5805505f, 2.914749f);
lat.setText(String.valueOf(latitude));
lon.setText(String.valueOf(longitude));
Geocoder geocoder = new Geocoder(AndroidGPSTrackingActivity.this, Locale.ENGLISH);
try
List<Address> addresses = geocoder.getFromLocation(latitude, longitude, 1);
if(addresses != null)
Address returnedAddress = addresses.get(0);
StringBuilder strReturnedAddress = new StringBuilder("Address:\n");
for(int i=0; i<returnedAddress.getMaxAddressLineIndex(); i++)
strReturnedAddress.append(returnedAddress.getAddressLine(i)).append("\n");
addr.setText(strReturnedAddress.toString());
else
addr.setText("No Address returned!");
catch (IOException e)
// TODO Auto-generated catch block
e.printStackTrace();
addr.setText("Can not get Address!");
// \n is for new line
//Toast.makeText(getApplicationContext(), "Your Location is - \nLat: " + latitude + "\nLong: " + longitude+" " +
//" diastance "+Distance, Toast.LENGTH_LONG).show();
else
// can't get location
// GPS or Network is not enabled
// Ask user to enable GPS/network in settings
gps.showSettingsAlert();
);
我在 logcat 中收到以下错误：
5-09 11:17:21.858 4501-4501/com.example.khaledsb.location E/AndroidRuntime: FATAL EXCEPTION: main
java.lang.IndexOutOfBoundsException: Invalid index 0, size is 0
at java.util.ArrayList.throwIndexOutOfBoundsException(ArrayList.java:251)
at java.util.ArrayList.get(ArrayList.java:304)
at com.example.khaledsb.location.AndroidGPSTrackingActivity$1.onClick(AndroidGPSTrackingActivity.java:79)
at android.view.View.performClick(View.java:4204)
at android.view.View$PerformClick.run(View.java:17355)
at android.os.Handler.handleCallback(Handler.java:725)
at android.os.Handler.dispatchMessage(Handler.java:92)
at android.os.Looper.loop(Looper.java:137)
at android.app.ActivityThread.main(ActivityThread.java:5041)
at java.lang.reflect.Method.invokeNative(Native Method)
at java.lang.reflect.Method.invoke(Method.java:511)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:793)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:560)
at dalvik.system.NativeStart.main(Native Method)
【问题讨论】：
查看以下链接，您可以在线获取完整地址：Get complete address from lat and long 您的问题有很多解决方案。你可以挖谷歌。 为什么要投反对票，你应该发表评论。 @AseshaGeorge 你在说谁？？ 投反对票的人 【参考方案1】：试试
Geocoder gcd = new Geocoder(this, Locale.getDefault());
List<Address> addresses = null;
try
addresses = gcd.getFromLocation(lat, lng, 1);
catch (IOException e)
e.printStackTrace();
if (addresses != null && addresses.size() > 0)
String locality = addresses.get(0).getLocality();
【讨论】：
要求使用try catch 请分享经纬度值 我明白了，我刚刚将 1,1 更改为纬度和经度 我不知道为什么有人拒绝对我的问题投票【参考方案2】：使用以下代码获取城市名称
Geocoder geocoder = new Geocoder(this);
try
List<Address>addresses = geocoder.getFromLocation(latitude,longitue,1);
if (geocoder.isPresent())
StringBuilder stringBuilder = new StringBuilder();
if (addresses.size()>0)
Address returnAddress = addresses.get(0);
String localityString = returnAddress.getLocality();
String name = returnAddress.getFeatureName();
String subLocality = returnAddress.getSubLocality();
String country = returnAddress.getCountryName();
String region_code = returnAddress.getCountryCode();
String zipcode = returnAddress.getPostalCode();
String state = returnAddress.getAdminArea();
else
catch (IOException e)
e.printStackTrace();
【讨论】：
【参考方案3】：试试这个代码，希望对你有帮助..
private static String getRegionName(double lati, double longi)
String regioName = "";
Geocoder gcd = new Geocoder(AppInstance, Locale.getDefault());
try
List<Address> addresses = gcd.getFromLocation(lati, longi, 1);
if (addresses.size() > 0)
regioName = addresses.get(0).getLocality();
catch (Exception e)
e.printStackTrace();
return regioName;
【讨论】：
【参考方案4】：尝试使用以下代码获取完整地址，
//Get address from Google api
//Log.e("Else","else");
try
JSONObject jsonObj = getJSONfromURL("http://maps.googleapis.com/maps/api/geocode/json?latlng=" + latitude + ","
+ longitude + "&sensor=true");
String Status = jsonObj.getString("status");
if (Status.equalsIgnoreCase("OK"))
JSONArray Results = jsonObj.getJSONArray("results");
JSONObject location = Results.getJSONObject(0);
finalAddress = location.getString("formatted_address");
catch (Exception e)
e.printStackTrace();
而不是，
geocoder.getFromLocation(latitude, longitude, 1);
【讨论】：
【参考方案5】：对于避免 ArrayIndexOutOfBound，您可以添加如下 if 条件：
if (addresses != null && addresses.size() > 0)
String locality = addresses.get(0).getLocality();
但是，当我看到您的代码时，您正在尝试使用 GeoCoder 按钮单击来获取地址，因此请避免这种情况，因为它是网络调用。您不应该在主线程中执行网络调用，它可能会阻塞 UI 或导致 ANR。您应该使用 AsyncTask/Thread 进行网络操作，以避免 UI 阻塞。
还要确保你已经添加了
INTERNET_PERMISSION
在您的清单文件中。
希望对你有帮助！
【讨论】：
【参考方案6】：错误表明 Array 列表为空，因此请确保初始化列表并检查它是否包含任何数据。
试试这个代码：
List<Address> addresses = new List<Address>;
addresses = geocoder.getFromLocation(latitude, longitude, 1);
【讨论】：
【参考方案7】：if(addresses != null)
String address = addresses.get(0).getAddressLine(0);
String city = addresses.get(0).getLocality();
String state = addresses.get(0).getAdminArea();
String country = addresses.get(0).getCountryName();
String postalCode = addresses.get(0).getPostalCode();
String knownName = addresses.get(0).getFeatureName();
这应该为您提供所需的信息。
【讨论】：
【参考方案8】：从 Location 对象获取的 latlong 坐标中获取位置地址。
@Override
public void onLocationChanged(Location location)
String longitude = "Longitude: " + location.getLongitude();
Log.v(TAG, longitude);
String latitude = "Latitude: " + location.getLatitude();
Log.v(TAG, latitude);
String address = null;
String knownName = null;
String city = null;
String state = null;
String country = null;
String postalCode = null;
Geocoder gcd = new Geocoder(MainActivity.this.getBaseContext(), Locale.getDefault());
List<Address> addresses;
try
addresses = gcd.getFromLocation(location.getLatitude(), location.getLongitude(), 1);
address = addresses.get(0).getAddressLine(0);
if (addresses.get(0).getFeatureName() != address)
knownName = addresses.get(0).getFeatureName();
city = addresses.get(0).getLocality();
state = addresses.get(0).getAdminArea();
country = addresses.get(0).getCountryName();
postalCode = addresses.get(0).getPostalCode();
catch (IOException e)
e.printStackTrace();
String foundAddress = knownName + ", " + address + ", " + city + ", " + state + ", " + country + "- " + postalCode + "\n";
myLocation.setText(foundAddress);
storeLocation();
【讨论】：
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