0

必须返回每个标题的描述我试图为每个工作创建路径但是返回全部而不是每个。有谁知道该怎么做?

javascript

$.getJSON("http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20xml%20where%20url%3D'http%3A%2F%2Frss.cnn.com%2Fservices%2Fpodcasting%2Fac360%2Frss.xml'%20AND%20itemPath%3D%22%2F%2Fchannel%22&format=json&diagnostics=true&callback=?", function (data) {

    // Load Titles patch Json
    console.log(data.query.results.channel.item);
    var titles = data.query.results.channel.item.map(function (item) {
        return item.title;

    });
    var urls = data.query.results.channel.item.map(function (item) {
        return item.origLink;

    });

    var descri = data.query.results.channel.item.map(function (item) {
        return item.description;

    });

    $(".description-podcast p").text(descri);

    console.log(titles);
    $(".container-list-podcast ul").append('<li>' + titles.join('</li><li>'));
    $(".container-list-podcast ul li").each(function (key, value) {
        var text = $(this).text();
        $(this).html('<a href="' + urls[key] + '">' + text + '</a>');
    });
    // Load Navigation Only Key
    a = $('.nav_holder li a').keynav(function () {
        return window.keyNavigationDisabled;
    });
});

jsfiddle

4

0 回答 0