jQuery在Chrome和Safari中工作,但不是Firefox或IE?

时间:2013-10-18 10:53:04

标签: javascript jquery google-chrome firefox cross-browser

我对一些JavaScript / jQuery代码感到困惑,我写的是在Webkit浏览器(Chrome和Safari)中完美运行,但在Firefox或IE中根本没有工作,我希望有人可以指出我的错误?

我正在做的是使用jQuery提取GeoRSS Feed,然后使用Leaflet在地图上绘制位置点。不知何故,使用Firefox或IE时没有绘制点? 这是有问题的页面:http://bit.ly/19N0I75

这是代码:

var map = L.mapbox.map('map', 'primitive.geography-class').setView([42, 22], 4);

var wordpressIcon = L.icon({
iconUrl: 'http://www.shifting-sands.com/wp-content/themes/shiftingsands/images/icons/wordpress.png',

iconSize:     [18, 18], // size of the icon
shadowSize:   [0, 0], // size of the shadow
iconAnchor:   [9, 9], // point of the icon which will correspond to marker's location
shadowAnchor: [0, 0],  // the same for the shadow
popupAnchor:  [0, 0] // point from which the popup should open relative to the iconAnchor
});

jQuery(document).ready(function($){
$.get("http://shifting-sands.com/feed/", function (data) {
var $xml = $(data);
var $i = 0;
$xml.find("item").each(function () {
    var $this = $(this),
        item = {
            title: $this.find("title").text(),
            linkurl: $this.find("link").text(),
            description: $this.find("description").text(),
            pubDate: $this.find("pubDate").text(),
            latitude: $this.find("lat").text(),
            longitude: $this.find("long").text()
        }

                lat = item.latitude;
                long = item.longitude;
                title = item.title;
                clickurl = item.linkurl;

                //Get the url for the image.
                var htmlString = '<h4><a href="' + clickurl + '" target="_blank">' + title + '</a></h4>';                       
                var contentString = '<div id="content">' + htmlString + '</div>';   

                //Create a new marker position using the Leaflet API.
                var rssmarker = L.marker([lat, long], {icon: wordpressIcon}).addTo(map);

                //Create a new info window using the Google Maps API

                rssmarker.bindPopup(contentString, {closeButton: true});


    $i++;
});
});
});

谢谢!

1 个答案:

答案 0 :(得分:0)

因为lat值是命名空间(geo:lat),所以你的find()没有任何价值。因此错误(,),两个空值,用逗号分隔。您必须将查询更改为:

latitude: $this.find("geo\\:lat").text()

双反斜杠逃脱冒号。