Swift3 阿拉莫菲尔邮政请求返回404 Swift格式

我将通过alamofire发布一个请求。我可以和邮递员一起寄出请求。响应返回true（201代码）。然而，当我尝试使用Alamofire时，返回404代码。怎么了

我的代码：

     let headers: HTTPHeaders = [
    "Authorization": "Basic xxxxxxxxxxxx",
    "content-type": "application/json"
     ]
        let parameters:[String:Any] = [
        "xxx":123,
        "yyy":"test",
        "zzz":"iphone"
        ]

    Alamofire.request(myUrl, method: .post, parameters: parameters, encoding: JSONEncoding.default, headers: headers).responseJSON { response -> Void in

        switch response.result {
        case .success:
            print(response.response?.statusCode)

            break
        case .failure(let error):

            print(error)
        }
    }

返回此消息：

success {
message = "No HTTP resource was found that matches the request URI 'https://xxxxx/PostErrorFeedBack'.";

---

}

试试这个，希望它能帮助你

let headers = [
    "Authorization": "Basic xxxxxxxxxxxx"
]
let parameters = [

]

Alamofire.request(.POST, "url", parameters: parameters, headers: headers, encoding: .JSON)
        .validate(contentType: ["application/json"])
        .responseJSON { response in
            if response.response?.statusCode == 200 {
                print("Success with JSON: \(response.result.value)")

                success(updatedUser)
            }
            else {
                let error = response.result.value as! NSDictionary
                let errorMessage = error.objectForKey("message") as! String
                print(errorMessage)
                failure(errorMessage)
            }


    }

---

您可以使用

Alamofire

尝试这种方法，因为这对我很有效

let headers = [
               "Accept": "application/json",
               "Authorization" : "Authorization: Bearer ", //if any
               "Cookie" : "Cookie" //if any
              ]

let parameterDict: NSDictionary = NSDictionary.init(objects: [nameTextField.text!, reportTextView.text!], forKeys: ["Name" as NSCopying,"Message" as NSCopying])

Alamofire.request("API",method: .post, parameters: parameterDict as? [String : AnyObject] , encoding:JSONEncoding.default, headers:headers) .responseJSON {  response in switch response.result {

     case .success(let JSON):
         print("Success with JSON: \(JSON)")
         let response = JSON as! NSDictionary

     case .failure(let error):
         print("Request failed with error: \(error)")


         }
    }

---

它是

内容类型

，而不是

内容类型

。同样，不会改变任何东西。您要点击的URL是什么？这是一个公共URL，还是你正在尝试点击你自己的本地服务器？根据消息，这似乎不是404，这是一个成功的请求。。。404到底在哪里？不幸的是，@Bikesh ThakurI没有得到这个错误“调用中的额外参数方法”，我在success updateuser中没有理解。这是怎么一回事？