如何调试jQuery Ajax请求？

My code is:

我的代码是：

var test = "it isn't working";
var response = $.ajax({
    type: 'GET',
    url: 'jquerydemo.php', //This is in the same site as the page calling this request, so it's not a same-domain error.
    success: function(){
            test = "it's working";
        },
    error: function(){
            alert("Error detected");
        }
 }).responseText;
alert(test);

I tested the status code and it came out to a 200 and the error function never goes off, but neither does the success function. Like I said in my comment, it isn't a same-origin policy error. It just stays saying "it isn't working". What's going on here?

我测试了状态代码，它出现了200并且错误函数永远不会消失，但成功函数也没有。就像我在评论中所说的那样，它不是同源政策错误。它只是说“它不起作用”。这里发生了什么？

5 个解决方案

#1

Try this:

尝试这个：

 error: function(jqXHR, textStatus, errorThrown) {
        console.log(JSON.stringify(jqXHR));
        console.log("AJAX error: " + textStatus + ' : ' + errorThrown);
 }

#2

Your ajax call is asynchronous. It has not completed yet when your alert at the end runs. Put an actual alert in the success function and you should see your result there.

你的ajax调用是异步的。当你的警报结束时，它还没有完成。在成功函数中添加实际警报，您应该在那里看到您的结果。

Remember that making the ajax call just starts the asynchronous ajax call and then the rest of your code continues to run. In the code example you posted, that means your alert(test) call runs right away before the ajax call has completed.

请记住，进行ajax调用只会启动异步ajax调用，然后其余代码继续运行。在您发布的代码示例中，这意味着您的警报（测试）调用在ajax调用完成之前立即运行。

You can ONLY examine the results of the ajax call from within the success handler itself.

您只能在成功处理程序本身内检查ajax调用的结果。

var test = "it isn't working";
var response = $.ajax({
    type: 'GET',
    url: 'jquerydemo.php', //This is in the same site as the page calling this request, so it's not a same-domain error.
    success: function(){
            alert("it's working");   // put this here and you will see it 
                                     // if the ajax call is successful
        },
    error: function(){
            alert("Error detected");
        }
 }).responseText;

#3

To debug these types of things, I find Firebug an indispensable tool. It will show you exactly the response from the server (500 error, 553 error, what have you). You can put break points in your Javascript code and debug it step by step. Firebug works on Firefox.

为了调试这些类型的东西，我发现Firebug是一个不可或缺的工具。它会准确显示服务器的响应（500错误，553错误，你有什么）。您可以在Javascript代码中添加断点并逐步调试它。 Firebug适用于Firefox。

For IE, you can use the Developer Tools feature which is similar to Firebug, specially on IE9 which seems more mature than previous versions of the Developer Tools for IE7 or IE8.

对于IE，您可以使用与Firebug类似的开发人员工具功能，特别是在IE9上，这似乎比以前版本的IE7或IE8的开发人员工具更成熟。

#4

Move that alert(test) from end into the success function of the ajax call. If it alerts it means code is working else it is not. you can only debug ajax call on its success return.

将该警报（测试）从结束移动到ajax调用的成功函数。如果它警告它意味着代码正在工作，否则它不是。你只能在成功返回时调试ajax调用。

var test = "it isn't working";
var response = $.ajax({
    type: 'GET',
    url: 'jquerydemo.php', 
    success: function(){
            test = "it's working";
            alert(test);   //It will alert when you ajax call returns successfully.
        },
    error: function(){
            alert("Error detected");
        }
 }).responseText;

Hope this helps.

希望这可以帮助。

#5

You can make it like

你可以做到这一点

  var response = $.ajax({
    type: 'GET',
    url: 'jquerydemo.php', 
    success: function(){
           alert("it's working");
        },
    error: function(){
            alert("Error detected");
        }
 }).responseText;

This will work....

这将工作....

#1