小程序中的地图定位不准确，是wgs84标准转火星标准的问题吧，之前用后台写能解决这个问题，现在js怎么解决这个转换呢，有没有在小程序里解决该问题的 以前程序里的代码是： public static final String BAIDU_LBS_TYPE = "bd09ll";     public static double pi = 3.1415926535897932384626;     public static double a = 6378245.0;     public static double ee = 0.00669342162296594323;     /**      * 84 to 火星坐标系 (GCJ-02) World Geodetic System ==> Mars Geodetic System      *      * @param lat      * @param lon      * @return      */     public static Gps gps84_To_Gcj02(double lat, double lon) {         if (outOfChina(lat, lon)) {             return null;         }         double dLat = transformLat(lon - 105.0, lat - 35.0);         double dLon = transformLon(lon - 105.0, lat - 35.0);         double radLat = lat / 180.0 * pi;         double magic = Math.sin(radLat);         magic = 1 - ee * magic * magic;         double sqrtMagic = Math.sqrt(magic);         dLat = (dLat * 180.0) / ((a * (1 - ee)) / (magic * sqrtMagic) * pi);         dLon = (dLon * 180.0) / (a / sqrtMagic * Math.cos(radLat) * pi);         double mgLat = lat + dLat;         double mgLon = lon + dLon;         return new Gps(mgLat, mgLon);     }

小程序里怎么把xml对象转成json 这些代码显然是用不了 function xmlToJson(xml) {// Create the return object var obj = {};if (xml.nodeType == 1) { // element // do attributes if (xml.attributes.length > 0) { obj["@attributes"] = {}; for (var j = 0; j < xml.attributes.length; j++) { var attribute = xml.attributes.item(j); obj["@attributes"][attribute.nodeName] = attribute.nodeValue; } } } else if (xml.nodeType == 3) { // text obj = xml.nodeValue; }// do children if (xml.hasChildNodes()) { for (var i = 0; i < xml.childNodes.length; i++) { var item = xml.childNodes.item(i); var nodeName = item.nodeName; if (typeof (obj[nodeName]) == "undefined") { obj[nodeName] = xmlToJson(item); } else { if (typeof (obj[nodeName].length) == "undefined") { var old = obj[nodeName]; obj[nodeName] = []; obj[nodeName].push(old); } obj[nodeName].push(xmlToJson(item)); } } } return obj; };