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:
登录
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。
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)
要强>
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"
要强>
let url = "http://192.169.201.32:9000/users/authenticate"
let url = "http://192.169.201.32:9000/patient/signUp"