Question

我在我的一个HTML文件中使用以下代码

var queryURL = encodeURI(yahooUrl + loc + appId);
alert(queryURL);

$.getJSON(queryURL, function(data){
    alert('inside getJSON')
    alert(data);
    var items = [];
    $.each(data, function(key, value){
        items.push('<li id="' + key + '">' + value + '</li>');
    });
    $('<ul/>', {
        'class': 'my-new-list',
        html: items.join('')
    }).appendTo('body');
});`

其中queryURL是一个大查询，如果我从浏览器的地址栏加载，我会得到一个包含JSON对象的文件。但是下面的代码不起作用，整个JSON对象显示在Firefox的错误控制台上，错误“无效标签”。我在查询字符串的末尾添加了&callback=?，如SO中的一些答案所述。

任何人都可以建议我做错了吗？

修改：for

queryURL =“http://where.yahooapis.com/geocode?location=107,South%20Market,San%20Jose,San%20Fransico,Leusina,USA,&flags=J&appid=dj0yJmk9SUk0NkdORm9qM2FyJmQ9WVdrOU1tVnFUVzlVTm5NbWNHbzlORFl4TnpZME5UWXkmcz1jb25zdW1lcnNlY3JldCZ4PWE1&callback= ？“

我收到以下错误：

Error: invalid label
Source File: http://where.yahooapis.com/geocode?location=107,South%20Market,San%20Jose,San%20Fransico,Leusina,USA,&flags=J&appid=dj0yJmk9SUk0NkdORm9qM2FyJmQ9WVdrOU1tVnFUVzlVTm5NbWNHbzlORFl4TnpZME5UWXkmcz1jb25zdW1lcnNlY3JldCZ4PWE1&callback=jQuery16404719878257064011_1316606312366&_=1316608283354
Line: 1, Column: 1

源代码：

{"ResultSet":{"version":"1.0","Error":0,"ErrorMessage":"No error","Locale":"us_US","Quality":87,"Found":1,"Results":[{"quality":39,"latitude":"37.336849","longitude":"-121.847710","offsetlat":"37.338470","offsetlon":"-121.885788","radius":34800,"name":"","line1":"","line2":"San Jose, CA","line3":"","line4":"United States","house":"","street":"","xstreet":"","unittype":"","unit":"","postal":"","neighborhood":"","city":"San Jose","county":"Santa Clara County","state":"California","country":"United States","countrycode":"US","statecode":"CA","countycode":"","uzip":"","hash":"","woeid":2488042,"woetype":7}]}}

Answer 1

这可能是因为jQuery自动切换到使用JSONP（因为它是跨域请求）和Yahoo apparently doesn't use JSONP but regular JSON。您是否尝试过使用$.ajax()的好dataType:"JSON"？

使用$ .ajax ：

        $.ajax({
            url: queryURL,
            dataType: "JSON",
            success: function(data){
              alert('inside getJSON')
              alert(data);
              var items = [];
              $.each(data, function(key, value){
                  items.push('<li id="' + key + '">' + value + '</li>');
              });
              $('<ul/>', {
                  'class': 'my-new-list',
                  html: items.join('')
              }).appendTo('body');
            }
        });

让我在这里非常好，因为我有一个可怕的日子：Working example

如何使用JavaScript从URL获取JSON？

1 个答案: