在回调函数中访问jQuery Ajax请求的URL

Question

有没有办法可以看到我用jQuery执行Ajax请求时请求的URL？

例如,

var some_data_object = { ...all sorts of junk... }
$.get('/someurl.php',some_data_object, function(data, textStatus, jqXHR) {
   var real_url = ? # <-- How do I get this
})

如何访问jQuery实际用于发出请求的URL？也许某些方法/属性jqHXR？我在文档中找不到它.

谢谢.

Answer 1

在success方法中设置一个断点,然后观察

this.url

是请求的真实网址.

Answer 2

这是一个可能的解决方案:

1. 通过实现beforeSend回调函数,在将ajax调用发送到服务器之前捕获它.
2. 保存网址和数据

3. 在您生成的错误消息中报告它.

   var url = "";
   
   $.ajax({
     url: "/Product/AddInventoryCount",
     data: { productId: productId, trxDate: trxDate, description: description, reference: reference, qtyInCount: qtyInCount }, //encodeURIComponent(productName)
     type: 'POST',
     cache: false,
     beforeSend: function (jqXHR, settings) {
       url = settings.url + "?" + settings.data;
     },
     success: function (r) {
       //Whatever            
     },
     error: function (jqXHR, textStatus, errorThrown) {
       handleError(jqXHR, textStatus, errorThrown, url);
     }
   });
   
   function handleError(jqXHR, textStatus, errorThrown, url) {
      //Whatever
   }
   

Answer 3

使用$.ajaxPrefilter:

// Make sure we can access the original request URL from any jqXHR objects
$.ajaxPrefilter(function(options, originalOptions, jqXHR) {
  jqXHR.originalRequestOptions = originalOptions;
});

$.get(
  'http://www.asdf.asdf'
).fail(function(jqXHR){
  console.log(jqXHR.originalRequestOptions);
  // -> Object {url: "http://www.asdf.asdf", type: "get", dataType: undefined, data: undefined, success: undefined}
});

http://api.jquery.com/jQuery.ajaxPrefilter/