如何调试jQuery Ajax请求？

Question

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);

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.它只是停留说"它不起作用".这里发生了什么？

Answer 1

试试这个:

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

Answer 2

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.

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alert(test)呼叫在ajax调用完成之前立即运行.

您只能从成功处理程序本身中检查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;