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從Android中沒有城市的地址獲取位置(緯度/經度)

[英]Get location (latitude / longitude) from address without city in Android

我試圖找到沒有城市的街道地址(緯度/經度) - 例如“123 Main St.” - 最接近當前位置。 此功能內置於Google Maps應用程序以及iOS地圖api中,因此令人驚訝地發現它缺少Android - 例如調用Geocoder.getFromLocation()並讓平台插入參考點。 我嘗試了幾種解決方案,以下是最好的,但仍然感覺自卑。

我使用左下角和右上角坐標調用Geocoder.getFromLocationName()。 呼叫從當前位置周圍的10kmx10km區域開始,並重復(30x30,100x100,然后沒有邊界框參數),直到返回一些地址。 返回多個地址時,計算並使用最接近的地址:

更新:這種方法似乎對於在邊界之外容易找到的地址效率低下。 例如,從紐約西海岸搜索“紐約,紐約”或“波士頓” - 需要3個有界和1個無限制的Geocoder.getFromLocation()調用。 然而,毫無疑問,在第一次調用時,紐約和波士頓會返回正確的lat / lng,CA中有最嚴格的界限。 谷歌很聰明,無視我們的界限。 這可能會給某些人帶來問題,但這種方法很有用。

package com.puurbuy.android;

import java.io.IOException;
import java.util.List;

import android.content.Context;
import android.location.Address;
import android.location.Geocoder;
import android.location.Location;
import android.os.AsyncTask;
import android.util.Log;

public class GeocoderRunner extends AsyncTask<String, Void, Address> {
    final static double LON_DEG_PER_KM = 0.012682308180089;
    final static double LAT_DEG_PER_KM =0.009009009009009;
    final static double[] SEARCH_RANGES = {10, 50,800,-1}; //city, region, state, everywhere

private Context mContext;
private GeocoderListener mListener;
private Location mLocation;

public GeocoderRunner(Context context, Location location,
        GeocoderListener addressLookupListener) {
    mContext = context;
    mLocation = location;
    mListener = addressLookupListener;
}
@Override
protected Address doInBackground(String... params) {
    Geocoder geocoder = new Geocoder(mContext);
    List<Address> addresses = null;

    //reference location TODO handle null 
    double lat = mLocation.getLatitude();
    double lon = mLocation.getLongitude();
    int i = 0;
    try {
        //loop through SEARCH_RANGES until addresses are returned
        do{
            //if range is -1, call  getFromLocationName() without bounding box
            if(SEARCH_RANGES[i] != -1){

                //calculate bounding box
                double lowerLeftLatitude =  translateLat(lat,-SEARCH_RANGES[i]);
                double lowerLeftLongitude = translateLon(lon,SEARCH_RANGES[i]);
                double upperRightLatitude = translateLat(lat,SEARCH_RANGES[i]);
                double upperRightLongitude = translateLon(lon,-SEARCH_RANGES[i]);

                addresses = geocoder.getFromLocationName(params[0], 5, lowerLeftLatitude, lowerLeftLongitude, upperRightLatitude, upperRightLongitude);

            } else {
                //last resort, try unbounded call with 20 result
                addresses = geocoder.getFromLocationName(params[0], 20);    
            }
            i++;

        }while((addresses == null || addresses.size() == 0) && i < SEARCH_RANGES.length );

    } catch (IOException e) {
        Log.i(this.getClass().getSimpleName(),"Gecoder lookup failed! " +e.getMessage());
    }



    if(addresses == null ||addresses.size() == 0)
        return null;

    //If multiple addresses were returned, find the closest
    if(addresses.size() > 1){
        Address closest = null;
        for(Address address: addresses){
            if(closest == null)
                closest = address;
            else
                closest = getClosest(mLocation, closest,address);//returns the address that is closest to mLocation
        }
        return closest;
    }else
        return addresses.get(0);

}

@Override
protected void onPostExecute(Address address) {
    if(address == null)
        mListener.lookupFailed();
    else
        mListener.addressReceived(address);

}

//Listener callback
public interface GeocoderListener{
    public void addressReceived(Address address);
    public void lookupFailed();
}

//HELPER Methods

private static double translateLat(double lat, double dx){
    if(lat > 0 )
        return (lat + dx*LAT_DEG_PER_KM);
    else
        return (lat - dx*LAT_DEG_PER_KM);
}

private static double translateLon(double lon, double dy){
    if(lon > 0 )
        return (lon + dy*LON_DEG_PER_KM);
    else
        return (lon - dy*LON_DEG_PER_KM);

}

private static Address getClosest(Location ref, Address address1, Address address2){
    double xO = ref.getLatitude();
    double yO = ref.getLongitude();
    double x1 = address1.getLatitude();
    double y1 = address1.getLongitude();
    double x2 = address2.getLatitude();
    double y2 = address2.getLongitude();
    double d1 = distance(xO,yO,x1,y1);
    double d2 = distance(xO,yO,x2,y2);
    if(d1 < d2)
        return address1;
    else
        return address2;

}

private static double distance(double x1, double y1, double x2, double y2){
    return Math.sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) );
}

}

也許這是最好的解決方案,但我想知道是否有辦法在一次通話中完成此操作。

您的代碼看起來太復雜了,這里有更簡單的方法:

    String searchPattern = "123 Main St." 
    LocationManager lm = (LocationManager) getSystemService(Context.LOCATION_SERVICE);
    //I use last known location, but here we can get real location
    Location lastKnownLocation = lm.getLastKnownLocation(LocationManager.GPS_PROVIDER);

    List<Address> addresses = null;
    try {
        //trying to get all possible addresses by search pattern
        addresses = (new Geocoder(this)).getFromLocationName(searchPattern, Integer.MAX_VALUE);
    } catch (IOException e) {
    }
    if (addresses == null || lastKnownLocation == null) {
        // location service unavailable or incorrect address
        // so returns null
        return null;
    }

    Address closest = null;
    float closestDistance = Float.MAX_VALUE;
    // look for address, closest to our location
    for (Address adr : addresses) {
        if (closest == null) {
            closest = adr;
        } else {
            float[] result = new float[1];
            Location.distanceBetween(lastKnownLocation.getLatitude(), lastKnownLocation.getLongitude(), adr.getLatitude(), adr.getLongitude(), result);
            float distance = result[0];
            if (distance < closestDistance) {
                closest = adr;
                closestDistance = distance;
            }
        }
    }
    return closest; //here can be null if we did not find any addresses by search pattern.

我嘗試了Jin35的建議並增加了Geocoder.getFromLocationName()的max_results,但結果並不理想。 首先,大的max_result,無限制的調用比我的模擬器上的5個結果,geocoord有界調用花了更長的時間(2.5x - 7x = 1 - 6秒)。 也許現實世界會更快,這個因素變得不那么重要了。

殺手是無論max_results是50還是100, 每次只有20個結果回來 似乎Google正在限制服務器端的結果。 最接近的“123 Main St”對我來說並不是那20個結果 - 從加利福尼亞州的Mt View進行測試,然后返回加利福尼亞州的Oakley。

除非Geocoder.getFromLocationName()之外的其他方法用於進行地址查找,或者更好的方式來使用邊界坐標,我將接受我自己的原始答案。

    getFromLocationName(String locationName, int maxResults, double lowerLeftLatitude, double lowerLeftLongitude, double upperRightLatitude, double upperRightLongitude)

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