php - 无法清除错误从类型 '(_, _, _) throws -> ()' 的抛出函数到非抛出函数类型的无效转换

Question

我收到标题中描述的错误。

Invalid conversion from throwing function of type '(_, _, _) throws -> ()' to non-throwing function type '(NSData?, NSURLResponse?, NSError?) -> Void'

在 session dataTaskWithRequest 之后。这是一个简单的用户注册到 PHP 和 mysql。在 Swift xcode 7.3 中。

 func displaymyalertmessage(userMessage:String)
    {
        var myAlert = UIAlertController(title: "Alert" , message:userMessage, preferredStyle: UIAlertControllerStyle.Alert); 
        let OkAction = UIAlertAction (title: "Ok", style: UIAlertActionStyle.Default, handler:nil);
        myAlert.addAction(OkAction);
        self.presentViewController(myAlert, animated:true, completion:nil);
    }

    // check for empty fields
    if(useremail!.isEmpty || username!.isEmpty ||
    userpassword!.isEmpty ||
        userrepeatpassword!.isEmpty){
        // Display alert message
            displaymyalertmessage("All fields are required");
        return; 
    }
        //Check if passwords match
        if(userpassword != userrepeatpassword)
        { 
            // display an alert message

            displaymyalertmessage("Passwords do not match");
            return;   
        } 
    // send data to server
    let myUrl = NSURL(string: "http://example/xxx.php")
    let request = NSMutableURLRequest(URL:myUrl!);
    var session = NSURLSession.sharedSession()
    request.HTTPMethod = "POST";

    let postString = "email=\(useremail)&password=\(userpassword)&username=\(username)"

    request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding);

在下一个任务中我收到错误 (_, _, _)

    do {
    let task = session.dataTaskWithRequest(request)
    { (data, response, error) in          
        // if error          
        if error != nil {
            print("error=\(error)")
            return
        }          
        var err: NSError?
        var json = try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers, error: &err) as? NSDictionary

        if let parseJSON = json {

            var resultvalue = parseJSON["status"] as? String
            printld("result: \(resultvalue)")

        var isuserregistered:Bool = false;
        if(resultvalue=="Success") { isuserregistered = true; }

        var messagetodisplay:String = parseJSON["message"] as String!;
        if(!isuserregistered)
        {
            messagetodisplay = parseJSON["message"] as String!;

        }

    dispatch_async(dispatch_get_main_queue(), {

    // Display alert message with confirmation

    var myAlert = UIAlertController(title: "Alert" , message: "Registration is successful. Thank You", preferredStyle: UIAlertControllerStyle.Alert);

    let OkAction = UIAlertAction (title: "Ok", style: UIAlertActionStyle.Default){ ACTION in
        self.dismissViewControllerAnimated(true, completion: nil);
    }

    myAlert.addAction(OkAction);
    self.presentViewController(myAlert, animated:true, completion:nil)

    });

        }    }
    task.resume()
    }

我已经尝试了很多东西，但我无法清除这个错误。

Answer 1

您需要将 try-catch 移动到回调 block 作用域中。此外，您不应通过引用将 err 传递给 JSONObjectWithData，因为它会引发错误。这应该有效:

let task = session.dataTaskWithRequest(request) { (data, response, error) in          
    // if error          
    if error != nil {
        print("error=\(error)")
        return
    }          

    var json: NSDictionary? = nil
    do {
       json = try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSDictionary
    } catch {
        print("error=\(error)")
    }
    //...