使用JSONP错误jQuery的Ajax请求 [英] jQuery ajax request using jsonp error

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本文介绍了使用JSONP错误jQuery的Ajax请求的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在写一个应用程序,我需要从其他服务器访问一些JSON数据在客户端。由于跨域问题我打算使用JSONP。 jQuery的可以让我这个用$ .getJSON()方法来做,但是,我没有办法告诉如果该方法失败(即服务器没有响应或东西)。所以我想尽了办法来获得使用$阿贾克斯,而不是JSON数据。但它不工作,我不知道什么尝试。 下面一个例子,显示我的问题:

I'm writing an app and I need to access some json data in the client side from another server. Because of the cross domain issue I plan on using jsonp. jQuery allows me to do this using the $.getJSON() method, however, I have no way to tell if the method has failed (i.e., the server is not responding or something). So I tried the approach to get the JSON data using $.ajax instead. But it's not working and I don't know what to try. Here an example showing my issue:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
 <html>
 <head>
      <meta http-equiv="Content-type" content="text/html; charset=utf-8">
      <title>TEST</title>
      <script type="text/javascript" src="scripts/jquery-1.5.1.min.js"></script>
      <script type="text/javascript">
           $(document).ready(function() {
        $('#button_detect').click(function(){
            var feedApiAjax = 'http://ajax.googleapis.com/ajax/services/feed/load?v=1.0&q=';
            var feedApiGetJSON = 'http://ajax.googleapis.com/ajax/services/feed/load?v=1.0&callback=?&q=';
            var feedUrl = 'http://www.engadget.com/rss.xml';

            $.ajax({
                url: feedApiAjax + feedUrl,
                datatype: 'jsonp',
                success: function(data) {
                    console.log('$.ajax() success');
                },
                error: function(xhr, testStatus, error) {
                    console.log('$.ajax() error');
                }
            });

            $.getJSON(
                feedApiGetJSON + feedUrl,
                function(data) {
                    console.log('$.getJSON success');
                });
        });
    });
      </script>
 </head>
 <body>
      <div id="button_detect">CLICK ME!!!!</div>
 </body>

如果您创建了这个code一个网页,点击点击我分区,你会看到$ .getJSON请求工作和$就一个没有了。我试图把/删除回调=? TG,用JSONP和JSON数据类型,但非这个工作。

If you create a web page with this code and click on the "Click Me" div you'll see that the $.getJSON request is working and the $.Ajax one is not. I've tried putting/removing the "callback=?" tg, used "jsonp" and "json" data types, but non of this worked.

这是什么可能我做任何想法错了?

Any idea on what might I be doing wrong?

干杯!

推荐答案

我以前也遇到过这样的错误,并且使用的的jQuery JSONP

I encountered this error before, and solved after using jquery-JSONP

[示例]

$.getJSON('http://server-url/Handler.ashx/?Callback=DocumentReadStatus',
  {
      userID: vuserID,
      documentID: vdocumentID
  },
  function(result) {
      if (result.readStatus == '1') {
          alert("ACCEPTED");
      }
      else if (result.readStatus == '0') {
          alert("NOT ACCEPTED");
      }
      else {
          alert(result.readStatus);
      }
  });

这篇关于使用JSONP错误jQuery的Ajax请求的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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