地图，测试当前位置是否在折线上或附近

问题描述

我正在使用google路线api为路线绘制多段线。有没有人有任何检查当前位置是否在多段线上/附近的例子？试图确定用户当前位置是否在该线的x米范围内，如果不是，我会提出新请求并重新绘制新路线。

干杯！

解决方案

以下是我的解决方案：只需将我创建的 bdccGeoDistanceAlgorithm 类添加到您的项目并使用 bdccGeoDistanceCheckWithRadius 方法检查您的当前位置是否在折线上（折线等于点的纬度列表）
您也可以获得距离从方法

类 bdccGeoDistanceAlgorithm

  import com.google.android.gms.maps.model.LatLng; 
 
 import java.util.List; 
 
 public class bdccGeoDistanceAlgorithm {
 
 //从GLatLng指向GPolyline或GPolygon的距离（以米为单位）poly 
 public static boolean bdccGeoDistanceCheckWithRadius（List< LatLng> poly，LatLng point ，int radius）
 {
 int i; 
 bdccGeo p = new bdccGeo（point.latitude，point.longitude）; （i = 0; i<（poly.size（） -  1）; i ++）
 {
 LatLng p1 = poly.get（i）; 
 
。 
 bdccGeo l1 = new bdccGeo（p1.latitude，p1.longitude）; 
 
 LatLng p2 = poly.get（i + 1）; 
 bdccGeo l2 = new bdccGeo（p2.latitude，p2.longitude）; 
 
 double distance = p.function_distanceToLineSegMtrs（l1，l2）; 
 
 if（距离< radius）
返回true; 
} 
返回false; 
} 
 
 
 // object 
 
 public static class bdccGeo 
 {
 public double lat; 
 public double lng; 
 
 public double x; 
 public double y; 
 public double z; 
 
 
 public bdccGeo（double lat，double lon）{
 this.lat = lat; 
 this.lng = lng; 
 
 double theta =（lon * Math.PI / 180.0）; 
 double rlat = function_bdccGeoGeocentricLatitude（lat * Math.PI / 180.0）; 
 double c = Math.cos（rlat）; 
 this.x = c * Math.cos（theta）; 
 this.y = c * Math.sin（theta）; 
 this.z = Math.sin（rlat）; 
} 
 
 //以米为单位返回此点与线段geo1-geo2 
 //的垂直距离的最小值，以及此点到线段的距离在geo1和geo2中结束
 public double function_distanceToLineSegMtrs（bdccGeo geo1，bdccGeo geo2）
 {
 
 //指向单位球体上方原点和法线到geo1平面，geo2 
 //可以是飞机的任一侧
 bdccGeo p2 = geo1.function_crossNormalize（geo2）; 
 
 //使用GC geo1 / geo2通过p传递给geo1 / geo2的GC法线的交点
 bdccGeo ip = function_bdccGeoGetIntersection（geo1，geo2，this，p2）; 
 
 //需要检查ip或其反对象在p1和p2之间
 double d = geo1.function_distance（geo2）; 
 double d1p = geo1.function_distance（ip）; 
 double d2p = geo2.function_distance（ip）; 
 //window.status = d +，+ d1p +，+ d2p; （（d> = d1p）&（d> = d2p））
返回function bbdccGeoRadiansToMeters（this.function_distance（ip））; 
 else 
 {
 ip = ip.function_antipode（）; 
 d1p = geo1.function_distance（ip）; 
 d2p = geo2.function_distance（ip）; （（d> = d1p）&（d> = d2p））
返回function_bdccGeoRadiansToMeters（this.function_distance（ip））; 
} 
。 
 else 
 return function_bdccGeoRadiansToMeters（Math.min（geo1.function_distance（this），geo2.function_distance（this）））; 
 
 
 //更多数学
 public bdccGeo function_crossNormalize（bdccGeo b）
 {
 double x =（this.y * bz） - （this .z * by）; 
 double y =（this.z * b.x） - （this.x * b.z）; 
 double z =（this.x * b.y） - （this.y * b.x）; 
 double L = Math.sqrt（（x * x）+（y * y）+（z * z））; 
 bdccGeo r = new bdccGeo（0,0）; 
 r.x = x / L; 
 r.y = y / L; 
 r.z = z / L; 
 
 return r; 
} 
 
 //返回由圆弧geo1到geo2和
 // geo3到geo4定义的两个很好的
 //圆圈的两个对极点。返回一个点作为Geo，使用.antipode获取另一个点
 public bdccGeo function_bdccGeoGetIntersection（bdccGeo geo1，bdccGeo geo2，bdccGeo geo3，bdccGeo geo4）
 {
 bdccGeo geoCross1 = geo1.function_crossNormalize （GEO2）; 
 bdccGeo geoCross2 = geo3.function_crossNormalize（geo4）; 
返回geoCross1.function_crossNormalize（geoCross2）; 
} 
 
 public double function_distance（bdccGeo v2）
 {
 return Math.atan2（v2.function_crossLength（this），v2.function_dot（this））; 
 
 
 //更多数学
 public double function_crossLength（bdccGeo b）
 {
 double x =（this.y * bz） - （this .z * by）; 
 double y =（this.z * b.x） - （this.x * b.z）; 
 double z =（this.x * b.y） - （this.y * b.x）; 
 return Math.sqrt（（x * x）+（y * y）+（z * z））; 
} 
 
 //数学
 public double function_dot（bdccGeo b）
 {
 return（（this.x * bx）+（this.y * by）+（this.z * bz））; 
} 
 
 //从Radians到Meters 
 public double function_bdccGeoRadiansToMeters（double rad）
 {
 return rad * 6378137.0; // WGS84赤道半径（以米为单位）
} 
 
 //指向这个点的世界另一端
 public bdccGeo function_antipode（）
 {
 return this.function_scale（-1.0）; 
} 
 
 //更多数学
 public bdccGeo function_scale（double s）
 {
 bdccGeo r = new bdccGeo（0,0）; 
 r.x = this.x * s; 
 r.y = this.y * s; 
 r.z = this.z * s; 
 return r; 
} 
 
 //将地理转换为以地心为中心的纬度（弧度）。 
 public double function_bdccGeoGeocentricLatitude（double geographicLatitude）
 {
 double flattening = 1.0 / 298.257223563; // WGS84 
 double f =（1.0  -  flattening）*（1.0  -  flattening）; 
 return Math.atan（（Math.tan（geographicLatitude）* f））; 
} 
} 
  

}

I'm using google directions api to draw a polyline for a route. Does anyone have any examples of checking if current location is on/near a polyline? Trying to determine if users current location is within x meters of that line and if not i'll make a new request and redraw a new route.

Cheers!

解决方案

Here is my solution: just add the bdccGeoDistanceAlgorithm class I have created to your project and use bdccGeoDistanceCheckWithRadius method to check if your current location is on or near polyline (polyline equals to a list of LatLng of points)Your can also get the distance from the method

Class bdccGeoDistanceAlgorithm

import com.google.android.gms.maps.model.LatLng;

import java.util.List;

public class bdccGeoDistanceAlgorithm {

// distance in meters from GLatLng point to GPolyline or GPolygon poly
public static boolean bdccGeoDistanceCheckWithRadius(List<LatLng> poly, LatLng point, int radius)
{
    int i;
    bdccGeo p = new bdccGeo(point.latitude,point.longitude);

    for(i=0; i < (poly.size()-1) ; i++)
    {
        LatLng p1 = poly.get(i);
        bdccGeo l1 = new bdccGeo(p1.latitude,p1.longitude);

        LatLng p2 = poly.get(i+1);
        bdccGeo l2 = new bdccGeo(p2.latitude,p2.longitude);

        double distance = p.function_distanceToLineSegMtrs(l1, l2);

        if(distance < radius)
            return true;
    }
    return false;
}


// object

public static class bdccGeo
{
    public double lat;
    public double lng;

    public double x;
    public double y;
    public double z;


    public bdccGeo(double lat, double lon) {
        this.lat = lat;
        this.lng = lng;

        double theta = (lon * Math.PI / 180.0);
        double rlat = function_bdccGeoGeocentricLatitude(lat * Math.PI / 180.0);
        double c = Math.cos(rlat);
        this.x = c * Math.cos(theta);
        this.y = c * Math.sin(theta);
        this.z = Math.sin(rlat);
    }

    //returns in meters the minimum of the perpendicular distance of this point from the line segment geo1-geo2
    //and the distance from this point to the line segment ends in geo1 and geo2
    public double function_distanceToLineSegMtrs(bdccGeo geo1,bdccGeo geo2)
    {

        //point on unit sphere above origin and normal to plane of geo1,geo2
        //could be either side of the plane
        bdccGeo p2 = geo1.function_crossNormalize(geo2);

        // intersection of GC normal to geo1/geo2 passing through p with GC geo1/geo2
        bdccGeo ip = function_bdccGeoGetIntersection(geo1,geo2,this,p2);

        //need to check that ip or its antipode is between p1 and p2
        double d = geo1.function_distance(geo2);
        double d1p = geo1.function_distance(ip);
        double d2p = geo2.function_distance(ip);
        //window.status = d + ", " + d1p + ", " + d2p;
        if ((d >= d1p) && (d >= d2p))
            return function_bdccGeoRadiansToMeters(this.function_distance(ip));
        else
        {
            ip = ip.function_antipode();
            d1p = geo1.function_distance(ip);
            d2p = geo2.function_distance(ip);
        }
        if ((d >= d1p) && (d >= d2p))
            return function_bdccGeoRadiansToMeters(this.function_distance(ip));
        else
            return function_bdccGeoRadiansToMeters(Math.min(geo1.function_distance(this),geo2.function_distance(this)));
    }

    // More Maths
    public bdccGeo function_crossNormalize(bdccGeo b)
    {
        double x = (this.y * b.z) - (this.z * b.y);
        double y = (this.z * b.x) - (this.x * b.z);
        double z = (this.x * b.y) - (this.y * b.x);
        double L = Math.sqrt((x * x) + (y * y) + (z * z));
        bdccGeo r = new bdccGeo(0,0);
        r.x = x / L;
        r.y = y / L;
        r.z = z / L;

        return r;
    }

    // Returns the two antipodal points of intersection of two great
    // circles defined by the arcs geo1 to geo2 and
    // geo3 to geo4. Returns a point as a Geo, use .antipode to get the other point
    public bdccGeo function_bdccGeoGetIntersection(bdccGeo geo1,bdccGeo  geo2, bdccGeo geo3,bdccGeo geo4)
    {
        bdccGeo geoCross1 = geo1.function_crossNormalize(geo2);
        bdccGeo geoCross2 = geo3.function_crossNormalize(geo4);
        return geoCross1.function_crossNormalize(geoCross2);
    }

    public double function_distance(bdccGeo v2)
    {
        return Math.atan2(v2.function_crossLength(this), v2.function_dot(this));
    }

    //More Maths
    public double function_crossLength(bdccGeo b)
    {
        double x = (this.y * b.z) - (this.z * b.y);
        double y = (this.z * b.x) - (this.x * b.z);
        double z = (this.x * b.y) - (this.y * b.x);
        return Math.sqrt((x * x) + (y * y) + (z * z));
    }

    //Maths
    public double function_dot(bdccGeo b)
    {
        return ((this.x * b.x) + (this.y * b.y) + (this.z * b.z));
    }

    //from Radians to Meters
    public double function_bdccGeoRadiansToMeters(double rad)
    {
        return rad * 6378137.0; // WGS84 Equatorial Radius in Meters
    }

    // point on opposite side of the world to this point
    public bdccGeo function_antipode()
    {
        return this.function_scale(-1.0);
    }

    //More Maths
    public bdccGeo function_scale(double s)
    {
        bdccGeo r = new bdccGeo(0,0);
        r.x = this.x * s;
        r.y = this.y * s;
        r.z = this.z * s;
        return r;
    }

    // Convert from geographic to geocentric latitude (radians).
    public double function_bdccGeoGeocentricLatitude(double geographicLatitude)
    {
        double flattening = 1.0 / 298.257223563;//WGS84
        double f = (1.0 - flattening) * (1.0 - flattening);
        return Math.atan((Math.tan(geographicLatitude) * f));
    }
}

}

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