jquery - Firefox 中出现无效标签错误

Question

当我从另一个域接收一些 json 内容时，我在 Firefox 中收到无效标签错误。我的 JS 代码是:

$.getJSON('http://www.upsidelearning.com/blog/blogapp/getBlogDetails.php?getBlogsList=1&jsoncallback=?', function(data){
     alert("Success");
      }); 

并且在 Chrome 错误控制台中收到以下错误:

(Warning)Resource interpreted as Script but transferred with MIME type text/html.
(Error) Uncaught SyntaxError: Unexpected token :

编辑:我的响应 json 是:

{"data":[{"title":"How Long Does It Take To Develop An Hour Of Elearning?","publishDate":"15 September 2011","author":"Abhijit Kadle","permalink":"http:\/\/www.upsidelearning.com\/blog\/index.php\/2011\/09\/15\/how-long-does-it-take-to-develop-an-hour-of-elearning\/","thumbnail":"http:\/\/www.upsidelearning.com\/blog\/blogapp\/images\/blogpost-icon.png","id":9197}]}

请帮助我。

Answer 1

I am getting Invalid label error in firefox while receiving some json content from another domain.

由于same origin policy限制 不允许向不同域发送 AJAX 请求。有两种解决方法:

1. 使用 JSONP。 getJSON() 中有一个关于它的特定部分。文档。仅当远程域支持时才有效。例如，而不是:

   {"data":[{"title":"How Long Does It Take To Develop An Hour Of Elearning?","publishDate":"15 September 2011","author":"Abhijit Kadle","permalink":"http:\/\/www.upsidelearning.com\/blog\/index.php\/2011\/09\/15\/how-long-does-it-take-to-develop-an-hour-of-elearning\/","thumbnail":"http:\/\/www.upsidelearning.com\/blog\/blogapp\/images\/blogpost-icon.png","id":9197}]}
   

   远程域必须能够返回:

   callback({"data":[{"title":"How Long Does It Take To Develop An Hour Of Elearning?","publishDate":"15 September 2011","author":"Abhijit Kadle","permalink":"http:\/\/www.upsidelearning.com\/blog\/index.php\/2011\/09\/15\/how-long-does-it-take-to-develop-an-hour-of-elearning\/","thumbnail":"http:\/\/www.upsidelearning.com\/blog\/blogapp\/images\/blogpost-icon.png","id":9197}]})
   

   客户端可以在其中设置callback姓名。

2. 如果 JSONP 不是一个选项，您可以在您的域上编写一个服务器端脚本，该脚本将充当您的域和远程域之间的桥梁，然后将 AJAX 请求发送到您的脚本。

<小时/>

更新:

经过大量评论后，您似乎正在尝试使用 YQL。这是 full working demo :

$.get('http://www.upsidelearning.com/blog/blogapp/getBlogDetails.php?getCategories=1&callback=?', function(result) {
    // you could parse the JSON like this:
    // var json = $.parseJSON(($(result.responseText).filter('p').text()));
    // alert(json.data[0].id);

    alert("Success");
});

请注意，您应该使用 $.get而不是$.getJSON .