26

我正在地图中的两点之间绘制路线。我以这种方式获得积分：

StringBuilder urlString = new StringBuilder();
    urlString.append("http://maps.googleapis.com/maps/api/directions/json");
    urlString.append("?origin=");// from
    urlString.append(Double.toString(src.latitude));
    urlString.append(",");
    urlString.append(Double.toString(src.longitude));
    urlString.append("&destination=");// to
    urlString.append(Double.toString(dest.latitude));
    urlString.append(",");
    urlString.append(Double.toString(dest.longitude));
    urlString.append("&sensor=false&mode=");
    if (tipo != null) {
        urlString.append(tipo);
    }
return urlString.toString;

我收到来自 Google 的响应，并得到 JSON：

resp = new JSONObject(builder.toString());
                Log.i("Location", "Contenido del kml: "+resp);
                JSONArray routeObject = resp.getJSONArray("routes");
                JSONObject routes = routeObject.getJSONObject(0);
                JSONObject overviewPolylines = routes
                        .getJSONObject("overview_polyline");
                String encodedString = overviewPolylines.getString("points");
                ArrayList<LatLng> puntos=decodePoly(encodedString);

我从谷歌得到的回复：（JSON 格式）：

Response: {"status":"OK","routes":[{"waypoint_order":[],"summary":"R-3 and A-3","bounds":{"southwest":{"lng":-3.676540000000001,"lat":40.00040000000001},"northeast":{"lng":-2.99933,"lat":40.43357}},"legs":[{"duration":{"value":3267,"text":"54 mins"},"distance":{"value":85039,"text":"85.0 km"},"end_location":{"lng":-2.99933,"lat":40.00040000000001},"start_address":"Calle del General Díaz Porlier, 91, 28006 Madrid, Spain","end_address":"Camino Veinticuatro, 2, 16400 Tarancón, Cuenca, Spain","start_location":{"lng":-3.676540000000001,"lat":40.43331000000001},"via_waypoint":[],"steps":[{"html_instructions":"Head <b>north<\/b> on <b>Calle del Gral Díaz Porlier<\/b> toward <b>Calle de Maldonado<\/b>","duration":{"value":6,"text":"1 min"},"distance":{"value":29,"text":"29 m"},"end_location":{"lng":-3.676520000000001,"lat":40.43357},"polyline":{"points":"ecxuFjamUs@C"},"travel_mode":"DRIVING","start_location":{"lng":-3.676540000000001,"lat":40.43331000000001}},{"html_instructions":"Take the 1st <b>right<\/b> onto <b>Calle de Maldonado<\/b>","duration":{"value":62,"text":"1 min"},"distance":{"value":266,"text":"0.3 km"},"end_location":{"lng":-3.6734,"lat":40.43345},"polyline":{"points":"ydxuFfamUPkG@a@JuGGk@"},"travel_mode":"DRIVING","start_location":{"lng":-3.676520000000001,"lat":40.43357}},{"html_instructions":"Take the 2nd <b>right<\/b> onto <b>Calle de Fco. Silvela<\/b>","duration":{"value":57,"text":"1 min"},"distance":{"value":245,"text":"0.2 km"},"end_location":{"lng":-3.671830000000001,"lat":40.4316},"polyline":{"points":"adxuFvmlURQl@e@zAoATSTMf@]pBeBLK"},"travel_mode":"DRIVING","start_location":{"lng":-3.6734,"lat":40.43345}},{"html_instructions":"Slight <b>right<\/b> onto <b>Calle de Francisco Silvela<\/b>","duration":{"value":51,"text":"1 min"},"distance":{"value":437,"text":"0.4 km"},"end_location":{"lng":-3.66913,"lat":40.42827},"polyline":{"points":"oxwuF|clUJADAFCbBkAjAeA`BoA^[XUPMLMFEnAeALIdBuAZUPMXS"},"travel_mode":"DRIVING","start_location":{"lng":-3.671830000000001,"lat":40.4316}},{"html_instructions":"Turn <b>right<\/b> onto <b>Pl. de Manuel Becerra<\/b>","duration":{"value":32,"text":"1 min"},"distance":{"value":154,"text":"0.2 km"},"end_location":{"lng":-3.668380000000001,"lat":40.42827},"polyline":{"points":"ucwuF`skU@?@@@@@?@@@?FBD?F@FAFAFADEDCFGDGDIBI@I@K@KAI?IAGAGAGCGCC?ACEEEIGGEICI?I?G@IBGD"},"travel_mode":"DRIVING","start_location":{"lng":-3.66913,"lat":40.42827}},{"html_instructions":"Turn <b>right<\/b> onto <b>Calle Alcala<\/b>","duration":{"value":61,"text":"1 min"},"distance":{"value":709,"text":"0.7 km"},"end_location":{"lng":-3.66107,"lat":40.43126000000001},"polyline":{"points":"ucwuFjnkUo@}A_AcCiEeLKYq@mBYq@[o@u@kB?KAQ]aAI[EQQu@Gc@QwAM_BEa@AU"},"travel_mode":"DRIVING","start_location":{"lng":-3.668380000000001,"lat":40.42827}},{"html_instructions":"Turn <b>right<\/b> onto the ramp to <b>M-30\/A-3\/A-4<\/b>","duration":{"value":44,"text":"1 min"},"distance":{"value":453,"text":"0.5 km"},"end_location":{"lng":-3.660580000000001,"lat":40.42720000000001},"polyline":{"points":"kvwuFt`jU\\CjIg@nBKD?rCOvDY"},"travel_mode":"DRIVING","start_location":{"lng":-3.66107,"lat":40.43126000000001}},{"html_instructions":"Continue straight","duration":{"value":45,"text":"1 min"},"distance":{"value":1009,"text":"1.0 km"},"end_location":{"lng":-3.659120000000001,"lat":40.41822000000001},"polyline":{"points":"_}vuFr}iUzKgBvCe@t@KvGm@hDUbAMPCZElAM|D_@^En@EjBEd@Df@DnAD"},"travel_mode":"DRIVING","start_location":{"lng":-3.660580000000001,"lat":40.42720000000001}},{"html_instructions":"Take exit <b>7B<\/b> to merge onto <b>M-23<\/b> toward <b>Vicálvaro\/R-3\/Valencia<\/b>","duration":{"value":117,"text":"2 mins"},"distance":{"value":2552,"text":"2.6 km"},"end_location":{"lng":-3.63392,"lat":40.41499},"polyline":{"points":"{duuFntiUFHB@^PDBFFFDDFDFBDDFBH@BBJ@F@H?F@H?FAH?FAF?FADAFAFCDADADEFCFEDEDEDEBEBEBGBE@G@E@G?G?EAG?GCGAGCEEGEEEEEEGEICIEIAKCKAK?K?MDgA?G?E?KJq@BODY@CBODW?C@I@AFe@F_@V{An@gED[@C?EHe@Ho@@G?A^{BjA}HF[tAiGzBwIdAcEVgApAuFr@sEF]Ju@DW?EDUHa@f@qEFo@f@iGXmHDgFCgGG{DM

最后，我以这种方式解码接收到的字符串，我在 Stack Overflow 的很多答案中都找到了它：

private ArrayList<LatLng> decodePoly(String encoded) {

    Log.i("Location", "String received: "+encoded);
    ArrayList<LatLng> poly = new ArrayList<LatLng>();
    int index = 0, len = encoded.length();
    int lat = 0, lng = 0;

    while (index < len) {
        int b, shift = 0, result = 0;
        do {
            b = encoded.charAt(index++) - 63;
            result |= (b & 0x1f) << shift;
            shift += 5;
        } while (b >= 0x20);
        int dlat = ((result & 1) != 0 ? ~(result >> 1) : (result >> 1));
        lat += dlat;

        shift = 0;
        result = 0;
        do {
            b = encoded.charAt(index++) - 63;
            result |= (b & 0x1f) << shift;
            shift += 5;
        } while (b >= 0x20);
        int dlng = ((result & 1) != 0 ? ~(result >> 1) : (result >> 1));
        lng += dlng;

        LatLng p = new LatLng((int) (((double) lat /1E5)* 1E6), (int) (((double) lng/1E5   * 1E6)));
        poly.add(p);
    }
   
   for(int i=0;i<poly.size();i++){
       Log.i("Location", "Point sent: Latitude: "+poly.get(i).latitude+" Longitude: "+poly.get(i).longitude);
   }
    return poly;
}

这是我认为错误的地方，因为我设置了一个 for 来查看正在解码的点，答案类似于：

04-10 13:33:54.578: I/Location(25065): Point sent: Latitude: 90.0 Longitude: -110.0
04-10 13:33:54.578: I/Location(25065): Point sent: Latitude: 90.0 Longitude: -120.0
04-10 13:33:54.578: I/Location(25065): Point sent: Latitude: 90.0 Longitude: 100.0
04-10 13:33:54.578: I/Location(25065): Point sent: Latitude: 90.0 Longitude: 120.0
04-10 13:33:54.578: I/Location(25065): Point sent: Latitude: 90.0 Longitude: 40.0

显然，这些点被错误地解析了。结果是，任何地方都没有画线。

我的问题是......有人知道如何解析谷歌的答案，并将其翻译成 LatLng 对象（不是 GeoPoints 对象）吗？

编辑：

这是即将被解码的字符串：

04-10 13:50:51.608: I/Location(25065): String to decode: gcxuFjamUq@CPkGLwHJ}@hCuBdEeDf@SrJuHnG{Ed@Fr@q@AmAm@k@e@DaIaSiEuLqAwJ~Mw@~ZwDtZuC`FKbCVjA`ANdAs@nBw@Py@]c@oAT}DhC{PlBcM~GaY|DyS`BmPRy]yBu\UsNvAqObAuEnC}KpAqIh@gK[}NcByWXoRt@}GlDeWvFcq@nFi`@~G}\|@aDdDoLn`@}gAfe@upA`JyV|CiN~@uId@{ZyBwVkBmOwB{^AsNt@iWb@qYcCyS_DkPmAaK{@gRN{RbCkWrFgUbHmPxLsP|JwIpPwN|G}JfE}I|C_KlCsNvAwSFaQj@i]jCmXbHg\pIgSpGaLnN}P`JsHfVaNbQkHtZyIrOsCnTaCx\aA`i@b@nSRz]K`J{AxJiDfNkJjU{WbKkJpGeEpVkPpEiEzMcQjQw]nJyM~IoJ`IoKnH{NrOaWtIeJdNgNrLcPjKqOhIuI~MeJxKgErR{Cn]EjQkBjHqBfJeE|\yTlI_JvEqJxHsS|Xuc@~IeR|EiRr@kEz@sNOoL_B}Lk@mCsFgKkCoGwAsG]}N`CuO`D}GjHoHtFwC|HgBhFg@fJ}BtHaF~D_FjLqSfGqE|FcBnGEnEv@xK`BzNeAxQfAlJ{@jIwEjE}FvBuF|B{NtCsNrBgEdHiIdGuIxDoFpGkFhDaBrHyBbF[pHPlDg@jEuBnCkCtFgKjEqEfIcGpC{D|GoNdAcExAkSjBgDrCsBpDa@vHLfSAfBk@nE_F|Im`@tO_ZxCyEzJmKtHyElG_ClZeF|j@sIlQaD|EuChGkHnK{MdIsMfPwa@bDeJpD_RpFsz@nA}IrCqIrYoa@|BeGlEiYpD}HbKoRfCyKdBwUjEmo@zBkKzDyJvCeQdB}LJqGmAqQJ_Gx@uFtHkO`ImMfMmXtFySpCkXRuN]aYVoT~@}LxB{NvD{OjJaYbFsLnCcEtLoMfJyXzIiXbIiL~D_EzQcMjDeF|AcFhA_M`BkXpC_KjCsDnDeClQcHvEgElFqJfF_Jp\mWzd@k]`WkWlVy`@nOcTz`@_s@rP_ZbNkQ~Tu^jJaUhCuIpHgUrPi\zd@uq@~McQh\e]vTaU~KcF~KoGhG{KzF{HzGmDbDwBfDiEzB}GbEmTrBsHpCuFxe@oq@xb@el@tDgC`ZeJnXkHrKmB~EeB~C{A`Ce@bCUrBoAnAuGSqIn@yEhDeOrCuQ|A}EvAwFdDgAfGSrHUJ\p@Bj@_A|Ao@tCsD|C}Rj@wAdCuW~Ds]VCJw@OUpEc_@jDKbDwB

如果使用 LatLng 格式是指创建 LatLng 对象的方式，则可以在 decodePoly 方法中看到。这些 latLng 对象被添加到 PolylineOptions 对象中，并且 this 添加到地图中，这样：

PolylineOption ruta=new PolylineOptions();
for(int i=0;i<puntos.size();i++){
ruta.add(new LatLng(puntos.get(i).latitude, puntos.get(i).longitude));                      
}//puntos is an array where the array returned by the decodePoly method are stored                  
ruta.color(Color.RED).width(7);                     
Polyline polygon=mapa.addPolyline(ruta);
4

7 回答 7

28

对于那些现在需要它的人来说，有一个开源库，其中包含许多关于 Google Maps Android API 的有用内容，包括折线的解码和编码。

在Android Maps Utils和Android Maps Util Github中查看。对于解码和编码使用：

PolyUtil.decode(String encodedPath);
PolyUtil.encode(List<LatLng> path);
于 2016-07-22T23:19:24.473 回答
27

我更改了我在 Google 中长时间搜索后找到的用于此的 decodePoly，现在路线已正确绘制。

http://wptrafficanalyzer.in/blog/route-between-two-locations-with-waypoints-in-google-map-android-api-v2/

改变

LatLng p = new LatLng((int) (((double) lat /1E5)* 1E6), (int) (((double) lng/1E5   * 1E6)));

为了

LatLng p = new LatLng((((double) lat / 1E5)),(((double) lng / 1E5)));

现在工作。

于 2013-04-10T15:38:09.037 回答
8

在 Javascript 中也一样

    function decodePolyline(encoded) {
        if (!encoded) {
            return [];
        }
        var poly = [];
        var index = 0, len = encoded.length;
        var lat = 0, lng = 0;

        while (index < len) {
            var b, shift = 0, result = 0;

            do {
                b = encoded.charCodeAt(index++) - 63;
                result = result | ((b & 0x1f) << shift);
                shift += 5;
            } while (b >= 0x20);

            var dlat = (result & 1) != 0 ? ~(result >> 1) : (result >> 1);
            lat += dlat;

            shift = 0;
            result = 0;

            do {
                b = encoded.charCodeAt(index++) - 63;
                result = result | ((b & 0x1f) << shift);
                shift += 5;
            } while (b >= 0x20);

            var dlng = (result & 1) != 0 ? ~(result >> 1) : (result >> 1);
            lng += dlng;

            var p = {
                latitude: lat / 1e5,
                longitude: lng / 1e5,
            };
            poly.push(p);
        }
        return poly;
    }
于 2016-02-02T11:55:10.987 回答
3

此方法对于解码折线很有用

 private List<LatLng> decodePoly(String encoded) {
    List<LatLng> poly = new ArrayList<>();
    int index = 0, len = encoded.length();
    int lat = 0, lng = 0;

    while (index < len) {
        int b, shift = 0, result = 0;
        do {
            b = encoded.charAt(index++) - 63;
            result |= (b & 0x1f) << shift;
            shift += 5;
        } while (b >= 0x20);
        int dlat = ((result & 1) != 0 ? ~(result >> 1) : (result >> 1));
        lat += dlat;

        shift = 0;
        result = 0;
        do {
            b = encoded.charAt(index++) - 63;
            result |= (b & 0x1f) << shift;
            shift += 5;
        } while (b >= 0x20);
        int dlng = ((result & 1) != 0 ? ~(result >> 1) : (result >> 1));
        lng += dlng;

        LatLng p = new LatLng((((double) lat / 1E5)),
                (((double) lng / 1E5)));
        poly.add(p);
    }

    return poly;
}

这就是我得到折线的方式

JSONObject poly = route.getJSONObject("overview_polyline");
String polyline = poly.getString("points");
polyLineList = decodePoly(polyline);
于 2018-01-11T06:24:37.033 回答
1

这是 iOS 中的实现，对于任何好奇的人，它完全基于 @Fustigador 的答案，但已转换为 iOS。请注意，我没有相应地格式化代码。我只是为了返回承诺的对象而添加了一些变量。

- (GMSPath *)decodedPolylinePathFromEncodedPolylineString:(NSString *)encodedPolylineString {


    NSString *decodedPolylineString = @"";
    GMSMutablePath *decodedPolylinePath = [GMSMutablePath new];
    CLLocationCoordinate2D decodedCoordinate;
    CLLocationDegrees latitude, longitude;

    int index = 0;
    NSUInteger len = encodedPolylineString.length;

    int lat = 0, lng = 0;

    while (index < len) {
        int b, shift = 0, result = 0;
        do {

            b = [encodedPolylineString characterAtIndex:index++] - 63;
            result |= (b & 0x1f) << shift;
            shift += 5;
        } while (b >= 0x20);
        int dlat = ((result & 1) != 0 ? ~(result >> 1) : (result >> 1));
        lat += dlat;

        shift = 0;
        result = 0;
        do {
            b = [encodedPolylineString characterAtIndex:index++] - 63;
            result |= (b & 0x1f) << shift;
            shift += 5;
        } while (b >= 0x20);
        int dlng = ((result & 1) != 0 ? ~(result >> 1) : (result >> 1));
        lng += dlng;

        latitude =  (((double) lat / 1E5));
        longitude = (((double) lng / 1E5));

        decodedCoordinate = CLLocationCoordinate2DMake(latitude, longitude);

        decodedPolylineString = [NSString stringWithFormat:@"%f%f", latitude, longitude];
        ;
        NSLog(@"%@",decodedPolylineString);

        [decodedPolylinePath addCoordinate:decodedCoordinate];
    }


    return decodedPolylinePath;
}
于 2016-01-03T06:05:06.170 回答
1

在 Pascal (Delphi) 中相同：

function DecodeLine(inputstring: String): String;
var
  lat, lon, lat_f, lon_f: Double;
  index: Integer;
  len: Integer;
  b: Integer;
  shift: Integer;
  decodeResult: Integer;
  encoded: String;
  dlat: Integer;
  dlng: Integer;
begin
  Result := '';

  encoded := inputstring;
  len := Length(encoded);
  index := 1;
  lat := 0;
  lon := 0;

  while (index <= len) do begin
    b := $20;
    shift := 0;
    decodeResult := 0;

    while (b >= $20) do begin
      b := ord(encoded[index]);
      b := b - 63;
      inc(index);
      decodeResult := decodeResult or (b and $1f) Shl shift;
      shift := shift + 5;
    end;

    if (decodeResult and 1) <> 0 then
      dlat := not (decodeResult shr 1)
    else
      dlat := decodeResult shr 1;
    lat := lat+dlat;

    shift := 0;
    decodeResult := 0;
    b := $20;
    while (b >= $20) do begin
      b := ord(encoded[index]);
      b := b-63;
      inc(index);
      decodeResult := decodeResult or (b and $1f) Shl shift;
      shift := shift + 5;
    end;

    if (decodeResult and 1) <> 0 then
      dlng := not (decodeResult shr 1)
    else
      dlng := decodeResult shr 1;
    lon := lon + dlng;

    { The coordinates of the point are used for our purposes }
    lon_f := lon/100000.0;
    lat_f := lat/100000.0;

    if Result <> '' then
      Result := Result + ', ';
    Result := Result + '(' + FloatToStr(lon_f) + ' - ' + FloatToStr(lat_f) + ')';
  end; //while
end;


procedure TForm1.Button1Click(Sender: TObject);
var
  str: String;
begin
  // Do stuff
  str := DecodeLine('ibgqGq}ycIMvM');
  ShowMessage(str);
end;
于 2017-06-01T07:23:45.000 回答
0 
private List<GeoPoint> decodePoly(String encoded){
     
     List<GeoPoint> poly = new ArrayList<GeoPoint>();
     int index = 0, len = encoded.length();
     int lat = 0, lng = 0;

     while (index < len){
            int b, shift = 0, result = 0;
            do{
            
               b = encoded.charAt(index++) - 63;
               result |= (b & 0x1f) << shift;
               shift += 5;
            }while(b >= 0x20);
            int dlat = ((result & 1) != 0 ?~(result >> 1) : (result >> 1));
            lat += dlat;

            shift = 0;
            result = 0;
            do{
               b = encoded.charAt(index++) - 63;
               result |= (b & 0x1f) << shift;
               shift += 5;
            }while(b >= 0x20);
            int dlng = ((result & 1) != 0 ?~(result >> 1) : (result >> 1));
            lng += dlng;

            GeoPoint p = new GeoPoint((int) (((double) lat / 1E5) * 1E6), (int) (((double) lng / 1E5) * 1E6));
            poly.add(p);
     } 
     return poly;
}
于 2021-05-17T06:36:41.037 回答