swift iOS 从异步获取值（post 数据）

Question

我在异步过程中获取值时遇到问题。

我的 class 是：

public class PostService: NSObject {

var result : String = String()

//this is my function


func request(var dataPost : String, var destination : String)->String{

    let request = NSMutableURLRequest(URL: NSURL(string: destination as String)!)
    request.HTTPMethod = "POST"
    let postString = dataPost
    request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding)
    let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
        data, response, error in

        if error != nil {
            println("error=\(error)")
            return
        }

        let responseString = NSString(data: data, encoding: NSUTF8StringEncoding)

        self.result = responseString as String!

    }

    task.resume()

//problem is here that result will fill after async

    return self.result       
}

}

// 我在这里使用这个函数

override func viewDidLoad() {
    super.viewDidLoad()
    var dataPost :String="Name=Ali"
    var dest : String =  "http://example.com/api/getUserName"
    var result = Post.request(dataPost, destination: dest)
    lbltext.text = result as String!
}

Answer 1

错误的方法

override func viewDidLoad() {  

   var result = Post.request(dataPost, destination: dest)// This is the method requesting web data from server asynchronously
    lbltext.text = result as String! // This is wrong, how can you expect data on the very next line , it will always be nil

你所做的，调用一个委托方法并将数据传递给你的视图控制器class，一旦在你的func request中完成下载，然后从中填充数据你的委托方法。

Answer 2

func request(var dataPost : String, var destination : String, successHandler: (response: String) -> Void)->String{
    let request = NSMutableURLRequest(URL: NSURL(string: destination as String)!)
    request.HTTPMethod = "POST"
    let postString = dataPost
    request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding)
    let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
        data, response, error in

        if error != nil {
            println("error=\(error)")
            return
        }

        let responseString = NSString(data: data, encoding: NSUTF8StringEncoding)

        //self.result = responseString as String!
        successHandler(responseString as String!);

    }

    task.resume()
}

override func viewDidLoad() {
    super.viewDidLoad()
    var dataPost :String="Name=Ali"
    var dest : String =  "http://example.com/api/getUserName"
    Post.request(dataPost, destination: dest, successHandler: {
        (response) in
        lbltext.text = response;
    });

}

类似下面的内容。基本上，您传递成功处理程序的回调并在回调函数内执行操作。

异步操作意味着，它们独立于代码的字节序列发生。这就是为什么当你进行异步调用时，你不能 return 一个值。代码的原始序列运行s 而异步调用运行s 在后台（或在另一个线程中）。这就是存在成功和错误处理程序的原因。它们允许您在异步请求完成时运行一段代码。

P.S请重新格式化您的问题并确保所有代码都正确缩进。

swift iOS 从异步获取值（post 数据）

swift iOS Get value from async (post data)

asynchronous

listener

dispatch

ios

swift