获取用于错误回调的AJAX响应主体

jQuery's AJAX error function has the following parameters:

jQuery的AJAX错误函数有以下参数:

error(XMLHttpRequest, textStatus, errorThrown)

What's the best cross-browser way to get the response body?

获得响应主体的最佳跨浏览器方式是什么?

Does this work (reliably in all browsers)?

这是否有效(在所有浏览器中都可靠)?

$.ajax({
  error: function(http) {
    alert(http.responseText);
  }
});

4 个解决方案

#1

-5

There is a hidden function that can extract the data from XHR istance:

有一个隐藏函数可以从XHR istance中提取数据:

var responseText = $.httpData(xhr)

If you pass "json" as a second parameter it will treat the response as a JSON string.

如果您将“json”作为第二个参数传递，它将把响应视为json字符串。

Note that you might get an error because there is no response (network problem for example). Make sure you cover that case as well. Also, I believe (not sure) that jQuery invokes the error handler if the server returns a 4xx or 5xx status.

注意，您可能会得到一个错误，因为没有响应(例如网络问题)。确保你也涵盖了这个案例。另外，我认为(不确定)如果服务器返回4xx或5xx状态，jQuery将调用错误处理程序。

#2

As of jQuery 1.4.1 you should use:

对于jQuery 1.4.1，您应该使用:

var json = JSON.parse(xhr.responseText);

See http://api.jquery.com/jQuery.parseJSON/.

见http://api.jquery.com/jQuery.parseJSON/。

#3

For a more recent and general answer (since jquery 1.5), I'd use the jqXHR object:

对于更近期和更一般的答案(自jquery 1.5以来)，我将使用jqXHR对象:

$.ajax(url).fail(function(jqXHR, textStatus, errorThrown) {
    alert(jqXHR.responseText);
})

Alternatively responseJSON can be used to get the response body already parsed

还可以使用responseJSON获取已经解析的响应体

$.ajax(url).fail(function(jqXHR, textStatus, errorThrown) {
    console.log(jqXHR.responseJSON);
})

#4

One straightforward usage example with jQuery:

jQuery一个简单的用法示例:

var url = '/';
$.get(url).then(
  function(response) {
      $("#result").html(response);
  },
  function(jqXHR) {
    $("#result").html('Error occurred: '+ jqXHR.statusText + ' ' + jqXHR.status);
  }
);

This should return the HTML of current website front page.

这应该返回当前网站首页的HTML。

Then try entering a nonsense URL that doesn't exist and see the error thrown. You should get "404 Not Found" from web server. Test it: JSFiddle here

然后尝试输入一个不存在的无意义URL并查看抛出的错误。您应该从web服务器获得“404 Not Found”。测试:JSFiddle这里

#1

-5