Ajax 请求返回 200 OK,但触发了错误事件而不是成功

Ajax request returns 200 OK, but an error event is fired instead of success(Ajax 请求返回 200 OK,但触发了错误事件而不是成功)
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问题描述

我已经在我的网站上实现了一个 Ajax 请求,并且我正在从网页调用端点.它总是返回 200 OK,但 jQuery 会执行错误事件.
我尝试了很多东西,但我无法弄清楚问题所在.我在下面添加我的代码:

I have implemented an Ajax request on my website, and I am calling the endpoint from a webpage. It always returns 200 OK, but jQuery executes the error event.
I tried a lot of things, but I could not figure out the problem. I am adding my code below:

var row = "1";
var json = "{'TwitterId':'" + row + "'}";
$.ajax({
    type: 'POST',
    url: 'Jqueryoperation.aspx?Operation=DeleteRow',
    contentType: 'application/json; charset=utf-8',
    data: json,
    dataType: 'json',
    cache: false,
    success: AjaxSucceeded,
    error: AjaxFailed
});
function AjaxSucceeded(result) {
    alert("hello");
    alert(result.d);
}
function AjaxFailed(result) {
    alert("hello1");
    alert(result.status + ' ' + result.statusText);
}

推荐答案

jQuery.ajax 尝试根据指定的 dataType 参数或服务器发送的 Content-Type 标头转换响应正文.如果转换失败(例如,如果 JSON/XML 无效),则会触发错误回调.

jQuery.ajax attempts to convert the response body depending on the specified dataType parameter or the Content-Type header sent by the server. If the conversion fails (e.g. if the JSON/XML is invalid), the error callback is fired.

您的 AJAX 代码包含:

Your AJAX code contains:

dataType: "json"

在这种情况下是 jQuery:

In this case jQuery:

将响应评估为 JSON 并返回一个 JavaScript 对象.[…]JSON数据被严格解析;任何格式错误的 JSON 都是被拒绝并引发解析错误.[…] 一个空洞的回应也是拒绝;服务器应该返回 null 或 {} 的响应.

Evaluates the response as JSON and returns a JavaScript object. […] The JSON data is parsed in a strict manner; any malformed JSON is rejected and a parse error is thrown. […] an empty response is also rejected; the server should return a response of null or {} instead.

您的服务器端代码返回具有 200 OK 状态的 HTML 片段.jQuery 期待有效的 JSON,因此触发错误回调,抱怨 parseerror.

Your server-side code returns HTML snippet with 200 OK status. jQuery was expecting valid JSON and therefore fires the error callback complaining about parseerror.

解决办法是从你的jQuery代码中去掉dataType参数,让服务端代码返回:

The solution is to remove the dataType parameter from your jQuery code and make the server-side code return:

Content-Type: application/javascript

alert("Record Deleted");

但我宁愿建议返回 JSON 响应并在成功回调中显示消息:

But I would rather suggest returning a JSON response and display the message inside the success callback:

Content-Type: application/json

{"message": "Record deleted"}

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