相关疑难解决方法(0)

我使用以下jQuery通过数据服务插入数据.事件虽然我得到状态响应201并且数据已成功插入我的数据库,但系统仍将其视为错误并给我"失败"警报？

我在这里错过了什么？

$.ajax({
    type: "POST",
    url: "http://localhost:49223/Form/WebDataService.svc/XMLForm(guid'1eaef3a0-d6df-45bf-a8f6-3e7292c0d77e')/XMLRecord/",
    data: JSON.stringify(record),
    contentType: "application/json; charset=utf-8",
    dataType: "json",
    success: function() {
        alert("Success");
    },
    error: function(xhr) {
        alert("fail");
    }
});

更新:

来自Fire Bug的调试消息:

Preferences

POST http://localhost:49223/Form/WebDataService.svc/X...ef3a0-d6df-45bf-a8f6-3e7292c0d77e%27)/XMLRecord/

POST http://localhost:49223/Form/WebDataService.svc/XMLForm(guid%271eaef3a0-d6df-45bf-a8f6-3e7292c0d77e%27)/XMLRecord/

jquery....min.js (line 127)
POST http://localhost:49223/Form/WebDataService.svc/X...ef3a0-d6df-45bf-a8f6-3e7292c0d77e%27)/XMLRecord/

POST http://localhost:49223/Form/WebDataService.svc/XMLForm(guid%271eaef3a0-d6df-45bf-a8f6-3e7292c0d77e%27)/XMLRecord/

201 Created 6.7s

POST http://localhost:49223/Form/WebDataService.svc/X...ef3a0-d6df-45bf-a8f6-3e7292c0d77e%27)/XMLRecord/

POST http://localhost:49223/Form/WebDataService.svc/XMLForm(guid%271eaef3a0-d6df-45bf-a8f6-3e7292c0d77e%27)/XMLRecord/

201 Created


get readyState 4

get responseText "{ "d" : {\r\n"__metadata"...\')/XMLForm"\r\n}\r\n}\r\n} }"

get responseXML null

get status 201

get statusText "Created"

我有一个ajax帖子:

$.post("/api/v1/payment_methods/create_credit_card", values)
.done (response) ->
  console.log("GOOD JOB")
.fail (response) ->
  console.log("Adas")

响应是201,然而,完成似乎没有捕获它,而是它将失败.我认为201会被认为是成功的,并且会被完成捕获.为什么它不起作用的任何想法？

注意:上面的代码是在coffeescript中,它并没有真正影响问题,但解释了我的语法