• 经纬度之间互相转换(高德转百度,高德转wgs84,百度转高德,百度转wgs84,wgs84转百度,wgs84转高德)


    /**
     * wgs84 地心坐标系 GPS原始知坐标体系。在中国,任何一个地图产品都不允许使用GPS坐标,据说是为了保密。GoogleEarth及GPS芯片使用
     * bd09 百度
     * gcj02 火星坐标系 高德、腾讯、Google中国地图使用
     */
    let locationUtils = {};
    
    var PI = 3.14159265358979324
    var x_pi = PI * 3000.0 / 180.0;
    function transformLat (x, y) {
      var ret = -100.0 + 2.0 * x + 3.0 * y + 0.2 * y * y + 0.1 * x * y + 0.2 * Math.sqrt(Math.abs(x))
      ret += (20.0 * Math.sin(6.0 * x * PI) + 20.0 * Math.sin(2.0 * x * PI)) * 2.0 / 3.0
      ret += (20.0 * Math.sin(y * PI) + 40.0 * Math.sin(y / 3.0 * PI)) * 2.0 / 3.0
      ret += (160.0 * Math.sin(y / 12.0 * PI) + 320 * Math.sin(y * PI / 30.0)) * 2.0 / 3.0
      return ret
    }
    function transformLon (x, y) {
      var ret = 300.0 + x + 2.0 * y + 0.1 * x * x + 0.1 * x * y + 0.1 * Math.sqrt(Math.abs(x))
      ret += (20.0 * Math.sin(6.0 * x * PI) + 20.0 * Math.sin(2.0 * x * PI)) * 2.0 / 3.0
      ret += (20.0 * Math.sin(x * PI) + 40.0 * Math.sin(x / 3.0 * PI)) * 2.0 / 3.0
      ret += (150.0 * Math.sin(x / 12.0 * PI) + 300.0 * Math.sin(x / 30.0 * PI)) * 2.0 / 3.0
      return ret
    }
    function locationDelta (lon, lat) {
      var a = 6378245.0; //  a: 卫星椭球坐标投影到平面地图坐标系的投影因子。
      var ee = 0.00669342162296594323; //  ee: 椭球的偏心率。
      var dLat = transformLat(lon - 105.0, lat - 35.0)
      var dLon = transformLon(lon - 105.0, lat - 35.0)
      var radLat = lat / 180.0 * PI
      var magic = Math.sin(radLat)
      magic = 1 - ee * magic * magic
      var sqrtMagic = Math.sqrt(magic)
      dLat = (dLat * 180.0) / ((a * (1 - ee)) / (magic * sqrtMagic) * PI)
      dLon = (dLon * 180.0) / (a / sqrtMagic * Math.cos(radLat) * PI)
      return [dLon, dLat]
    }
    
    locationUtils.gcj02Towgs84 = function (lng, lat) {
       var a = 6378245.0; //  a: 卫星椭球坐标投影到平面地图坐标系的投影因子。
      var ee = 0.00669342162296594323; //  ee: 椭球的偏心率。
      var lat = +lat;
      var lng = +lng;
      var dlat = transformLat(lng - 105.0, lat - 35.0);
      var dlng = transformLon(lng - 105.0, lat - 35.0);
      var radlat = lat / 180.0 * PI;
      var magic = Math.sin(radlat);
      magic = 1 - ee * magic * magic;
      var sqrtmagic = Math.sqrt(magic);
      dlat = (dlat * 180.0) / ((a * (1 - ee)) / (magic * sqrtmagic) * PI);
      dlng = (dlng * 180.0) / (a / sqrtmagic * Math.cos(radlat) * PI);
      var mglat = lat + dlat;
      var mglng = lng + dlng;
      return [lng * 2 - mglng, lat * 2 - mglat]
    }
    
    locationUtils.gcj02Tobd09 = function (lon, lat) {
      var x = lon
      var y = lat
      var z = Math.sqrt(x * x + y * y) + 0.00002 * Math.sin(y * x_pi)
      var theta = Math.atan2(y, x) + 0.000003 * Math.cos(x * x_pi)
      var bdLon = z * Math.cos(theta) + 0.0065
      var bdLat = z * Math.sin(theta) + 0.006
      return [bdLon, bdLat]
    }
    
    locationUtils.bd09Towgs84 = function (lng, lat) {
      let bd09ToGcj02Location = locationUtils.bd09Togcj02(lng, lat)
      let gcj02towgs84Location =  locationUtils.gcj02Towgs84(bd09ToGcj02Location[0], bd09ToGcj02Location[1])
    
      return gcj02towgs84Location
    }
      
    locationUtils.bd09Togcj02 = function(bdLon, bdLat) {
      var x = bdLon - 0.0065
      var y = bdLat - 0.006
      var z = Math.sqrt(x * x + y * y) - 0.00002 * Math.sin(y * x_pi);
      var theta = Math.atan2(y, x) - 0.000003 * Math.cos(x * x_pi);
      var gcjLon = z * Math.cos(theta);
      var gcjLat = z * Math.sin(theta)
      return [gcjLon, gcjLat]
    }
    
    locationUtils.wgs84Togcj02 = function (lon, lat) {
      var lat = +lat;
      var lon = +lon;
      var d = locationDelta(lon, lat);
      return [lon + d[0], lat + d[1]]
    }
    locationUtils.wgs84Tobd09 = function (lon, lat) {
      var lat = +lat;
      var lon = +lon;
      let gcj02 = locationUtils.wgs84Togcj02(lon, lat)
      return locationUtils.gcj02Tobd09(gcj02[0], gcj02[1])
    }
    export default locationUtils;
  • 相关阅读:
    Java的注释
    输入一个字符串并判断是否是对称字符串
    将数组的数据按照指定格式进行拼接并打印
    对输入的电话号码中间4位进行屏蔽
    统计输入的大小写字母个数
    svnadmin
    jenkins
    travis-ci
    metamodel
    KISSY
  • 原文地址:https://www.cnblogs.com/chailuG/p/12896497.html
Copyright © 2020-2023  润新知