Question

我最近在使用Alamofire时遇到了问题。

这是我的代码 LoginViewController.swift

class LoginViewController: UIViewController {

@IBOutlet weak var name: UITextField!
@IBOutlet weak var email: UITextField!
@IBOutlet weak var password: UITextField!

let baseApi = BaseApi()

override func viewDidLoad() {
    super.viewDidLoad()

}

@IBAction func loginButton(_ sender: UIButton) {

    let dict = ["name":name.text,"email":email.text, 
        "password":password.text]
    print("api succeed1")
    let result = baseApi.login(paras: dict as! [String : String])
    print("api succeed2")
    if result[0]["status"].string == "success" {
        print("api succeed3")
        present( UIStoryboard(name: "Main", bundle: 
        nil).instantiateViewController
        (withIdentifier:"TabBarController") 
        as! TabBarController, animated: true, completion: nil)
    }
}

BaseApi.swift

class BaseApi{

func login(paras : [String:String]) -> JSON {
    let url = URL(string: "http://127.0.0.1:8000/api/login")
    let result = baseApi(url: url!,paras: paras)
    print("BaseApi3333")
    return result
}

func baseApi(url : URL,paras : [String:String]) -> JSON {

    var json:JSON = []

    let toke = getToken()

    let parameters: Parameters = [
        "name": paras["name"]!,
        "email": paras["email"]!,
        "password": paras["password"]!
    ]

    let headers: HTTPHeaders = [
        "Authorization": "Basic "+toke,
        "Accept": "application/json"
    ]

    Alamofire.request(url, method: .post, parameters: parameters, encoding: JSONEncoding.default).responseJSON { response in
        switch response.result {
        case .success(let value):
            json = JSON(value)
            print("baseAp2222")
            print(json)
        case .failure(let error):
            print(error)
        }
    }
    print("baseApi111")
    print(json)
    return json
}
}

这是日志

api succeed1
baseApi111
[

]
BaseApi3333
api succeed2
baseAp2222
{
  "status_code" : 200,
  "status" : "success",
  "data" : {
    "token" : "xxxx"
  }
}

我的问题是为什么print（baseApi111）在打印之前出来（＆＃34; baseAp2222＆＃34;），我需要返回json，但看起来excute命令不对，所以返回json是nil，如何我应该解决这个问题吗？

Answer 1

你需要使用回调闭包来进行回调，你不能从api调用中返回这样的数据。

让我举个例子 - 以下方法是使用almofire调用api -

func fetchDataFromWebService<T: Mappable>(_ parameters: Dictionary<String , AnyObject>, closure:@escaping (_ response: T) -> Void){
    let url = getWebServiceUrl()

    //        let url = NSURL(string: getWebServiceUrl)
    print("parameters = \(parameters)")

    Alamofire.request(url, method: .get, parameters: parameters, headers: nil).responseJSON { (response:DataResponse<Any>) in
        switch(response.result) {
        case .success(_):
            if response.response?.statusCode == 200 || response.response?.statusCode == 201 {
                //   print(response.result.value)
                var user = Mapper<T>().map(JSONObject: response.result.value)
                //                    var user = Mapper<T>().map(response.result.value)
                if self.processSingleRecord() == true {

                    user = Mapper<T>().map(JSONObject: (response.result.value as! NSArray).object(at: 0))
                    //                        user = Mapper<T>().map(response.result.value?.objectAtIndex(0))
                }
                closure(user!)
                //                        print("user = ",user)
            }
            else if response.response?.statusCode == 0{
                //  print(self.DisplayNetworkAvailabilityMessage())
            }
            else  {
                if let _ = response.result.value as? Error {
                }
            }
            break

        case .failure(let error):
            debugPrint("getEvents error: \(error)")
            SVProgressHUD.dismiss()
            break

        }
    }

}

现在这里是调用这个方法 -

let anotherWebServiceHandler = DeviceTokenDataHandler.init()        anotherWebServiceHandler.fetchDataFromWebService(["":""], closure: { (response:SignUpResponse) -> Void in

    })

您需要了解sequential code execution - 和Closures

Swift使用Alamofire发出http请求

1 个答案: