webservice的输出没有使用alamofire

时间:2018-05-22 11:00:19

标签: ios json swift web-services alamofire

1)

我是使用alamofire的新手。这是我尝试使用alamofire进行Web服务检查我错了。我已经在loginviewcontroller.swift中创建了一个登录webservice,如下所示

let url="http://192.169.201.32:9000//users/authenticate"

        @IBAction func DoLogin(_ sender: AnyObject) {
            Alamofire.request(url, method: .post, parameters:["username":"andrews","password":"admin2"], encoding: URLEncoding.default)
                .responseJSON { response in
                    print("abcsign in")
                    print(response)
                    print("abcsign in3")
                    print(response.result)
                    //to get status code
                    if let status = response.response?.statusCode {
                        switch(status){
                        case 201:
                            print("example success")
                        default:
                            print("error with response status: \(status)")
                        }
                    }
                    //to get JSON return value
                    if let result = response.result.value {
                        let JSON = result as! NSDictionary
                        print("abcsign in 2")
                        print(JSON)

                    }

            }

            if(login_button.titleLabel?.text == "Logout")
            {
                let preferences = UserDefaults.standard
                preferences.removeObject(forKey: "session")

                LoginToDo()
            }
            else{
                login_now(username:username_input.text!, password: password_input.text!)
            }

        }

打印(响应)

  

FAILURE

     

responseSerializationFailed(Alamofire.AFError.ResponseSerializationFailu> reReason.jsonSerializationFailed(错误   Domain = NSCocoaErrorDomain Code = 3840"字符周围的值无效   2" UserInfo = {NSDebugDescription =字符2周围的值无效。}))

打印(response.result)

  

FAILURE

error with response status: 404

2)

第二个signUpviewcontroller.swift与signUp视图控制器连接。在signUpViewController.swift中,singUp webservice的代码如下

  let url="http://192.169.201.32:9000//patient/signUp"

    @IBAction func signUpButtonWasPressed(_ sender: Any) {
        Alamofire.request(url, method: .post, parameters:["dob":DateOfBirthTextFeild.text ,
                                                          "email":emailIdTextField.text ,
                                                          "firstName":FirstNameTextField.text ,
                                                          "gender":genderTextField.text ,
                                                          "lastName":LastNameTextField.text ,
                                                          "middleName":MiddleNameTextField.text ,
                                                          "password":passwordTextField.text ,    //password must be 8 char long
            "ssn":ssnTextField.text], encoding: URLEncoding.default)
            .responseJSON { response in
                print("abcsig up ")
                print(response)
                print("abcsign up 3")
                print(response.result)
                //to get status code
                if let status = response.response?.statusCode {
                    switch(status){
                    case 201:
                        print("example success")
                    default:
                        print("error with response status: \(status)")
                    }
                }
                //to get JSON return value
                if let result = response.result.value {
                    let JSON = result as! NSDictionary
                    print("abcsign up in 2")
                    print(JSON)

                }

        }
    }    

打印(回复)

  

FAILURE

     

responseSerializationFailed(Alamofire.AFError.ResponseSerializationFailu> reReason.jsonSerializationFailed(错误

     

Domain = NSCocoaErrorDomain Code = 3840"字符周围的值无效   2" UserInfo = {NSDebugDescription =字符2周围的值无效。}))

打印(response.result)

  

错误与响应状态:404

如何获得有效的json响应?

您可以从此链接https://drive.google.com/file/d/1Q__ydaQ7o0fKcFHdq6ymkxh52IEf7hMK/view?usp=sharing下载项目 在postman上,api显示了所需的json out put。

Sigin Up:

image1

image2

登录

image3

image4

提供正文中的参数。您可以添加将json url请求转换为post的.json参数。在下面的选项卡中,选择body,通过选择raw来将参数放在此处。

2 个答案:

答案 0 :(得分:0)

尝试此链接" http://192.169.201.32:9000/users/authenticate"

我认为你有一个额外的/

404 Not Found表示问题出在链接上 它找不到页面错误

答案 1 :(得分:0)

查看您分享的屏幕截图,我发现API调用中存在2个问题。

1)将编码从URLEncoding更改为JSONEncoding

2)通过避免请求中的双//来更新您的网址

执行以下更改:

    Alamofire.request(url,
                      method: .post,
                      parameters:["username":"andrews","password":"admin2"],
                      encoding: URLEncoding.default)

    Alamofire.request(url,
                      method: .post,
                      parameters:["username":"andrews","password":"admin2"],
                      encoding: JSONEncoding.default) 
    let parameters = ["dob":"16-08-2017",
                      "email":"ali45324@heurixtics.com",
                      "firstName":"abdul",
                      "gender":"male",
                      "lastName":"hasmi",
                      "middleName":"rauf",
                      "password":"12345678",
                      "ssn":"1235"]
    Alamofire.request(url,
                      method: .post,
                      parameters:parameters,
                      encoding: JSONEncoding.default)

您还需要更正您的请求网址:

let url = "http://192.169.201.32:9000//users/authenticate"
let url = "http://192.169.201.32:9000//patient/signUp"

let url = "http://192.169.201.32:9000/users/authenticate"
let url = "http://192.169.201.32:9000/patient/signUp"