Gxml.parse google maps migrating from v2 to v3

Question

Я хочу переместить карты Google с v2 на v3 в одном из приложений, которые я разрабатываю. Я видел эту ссылку Tips for Upgrading Gmaps v2 to v3 more quickly. Однако я не мог найти альтернативу для метода GXml.parse в API v2.

Я использую функцию downloadUrl, предоставляемую этим скриптом http://gmaps-samples-v3.googlecode.com/svn/trunk/xmlparsing/util.js.

function renderMap(mapControlFile) { 
      if (!mapControlFile || !xmlPath) { 
       downloadUrl(mapConfigFile, function(data,responseCode) { 

        try {errorTest = errorTest;} 
        catch(e) {errorTest = null;} 
        if (responseCode!=200 || errorTest=='yes') { 
         document.getElementById('mapCanvas').innerHTML = mapDataError; 
         document.getElementById('mapCanvas').innerHTML += '<p>Response Code: '+responseCode+'</p>'; 
         return; 
        } 
// The below line uses GXml which is now deprecated in v3 
        xmlDocument = GXml.parse(data); 
        //process config element -- restricted to 1 config element in the xml file 
        configElement = xmlDocument.documentElement.getElementsByTagName("config"); 
        //check for xml file parsing errors 
        if (configElement.length==0) { 
         document.getElementById('mapCanvas').innerHTML = mapDataError; //<div> tag in html document 
         return; 
        } 
        //try-catch handler for potential undeclared variable "auth" across all browsers (controls access during maintenance) 
        try {auth = auth;} 
        catch(e) {auth = null;} 
        //try-catch handler for potential undeclared variable "maintenanceTest" across all browsers (used for maintenance message testing) 
        try {maintenanceTest = maintenanceTest;} 
        catch(e) {maintenanceTest = null;} 
        //check if campus map system is offline for maintenance 
        if ((GXml.value(configElement[0].getElementsByTagName("offline")[0]) && auth==null) || maintenanceTest=='yes') { 
         document.getElementById('mapCanvas').innerHTML = GXml.value(configElement[0].getElementsByTagName("offlineMsg")[0]); //<div> tag in html document 
         return; 
        } 
        //initialize and load default map 
        xmlPath = GXml.value(configElement[0].getElementsByTagName("xmlPath")[0]); 
        overlayURL = GXml.value(configElement[0].getElementsByTagName("overlayURL")[0]); //set global variable 
        //create map and add controls (documentation at http://code.google.com/apis/maps/documentation/) 
        //map canvas is styled in ../css/campusmaps.css 
        //Google Maps API Version 2 
        map = new GMap2(document.getElementById("mapCanvas")); //<div> tag in html document 
        eval('map.setMapType('+GXml.value(configElement[0].getElementsByTagName("mapType")[0])+')'); 
        map.addControl(new GSmallZoomControl3D()); 
        zoomLevelMessage = GXml.value(configElement[0].getElementsByTagName("zoomLevelMessage")[0]); //set global variable 
        parseXml(xmlPath+GXml.value(configElement[0].getElementsByTagName("defaultMap")[0])); //function call to parse xml default map control file 
       }); //end downloadUrl() 
      } //end initialize and load default map 
      else { 
       parseXml(xmlPath+mapControlFile); //function call to parse xml map control files 
      } //end load all non-default maps 

    } //end renderMap() 

Answer 1

У вас есть варианты.

1. использовать стороннюю библиотеку, такую ​​как jquery.
2. использование GXml.js
3. написать XML разбора рутина, как в моей версии downloadxml.js

Вариант 2:

function parse(textDoc){ 
    try{ 
     if(typeof ActiveXObject!="undefined"&&typeof GetObject!="undefined"){ 
     var b=new ActiveXObject("Microsoft.XMLDOM"); 
     b.loadXML(textDoc); 
     return b; 
     }else if(typeof DOMParser!="undefined"){ 
     return(new DOMParser()).parseFromString(textDoc,"text/xml"); 
     }else{ 
     return Wb(textDoc); 
     } 
    } 
    catch(c){ 
     P.incompatible("xmlparse"); 
    } 
    try{ 
     return Wb(textDoc); 
    } 
    catch(c){ 
     P.incompatible("xmlparse"); 
     return document.createElement("div"); 
    } 
} 

Вариант 3:

function xmlParse(str) { 
    if (typeof ActiveXObject != 'undefined' && typeof GetObject != 'undefined') { 
    var doc = new ActiveXObject('Microsoft.XMLDOM'); 
    doc.loadXML(str); 
    return doc; 
    } 

    if (typeof DOMParser != 'undefined') { 
    return (new DOMParser()).parseFromString(str, 'text/xml'); 
    } 

    return createElement('div', null); 
}