alertView现在是alertController

Question

我试图在iOS中完成我的登录页面，在我将alertView更改为alertController之后，页面上写着“'UIAlertController‘类型的值没有成员'show'”。

我的代码是：

if ( username.isEqualToString("") || password.isEqualToString("") ) {

        var alertView:UIAlertController = UIAlertController()
        alertView.title = "Sign Up Failed!"
        alertView.message = "Please enter Username and Password"
        alertView.delegate = self
        alertView.addButtonWithTitle("OK")
        alertView.show()
    } else if ( !password.isEqual(confirm_password) ) {

        var alertView:UIAlertController = UIAlertController()
        alertView.title = "Sign Up Failed!"
        alertView.message = "Passwords doesn't Match"
        alertView.delegate = self
        alertView.addButtonWithTitle("OK")
        alertView.show()
    } else {

我发现了错误：

'UIAlertController‘没有成员’show‘和'UIAlertController’类型的值没有成员‘委托’等等。

此外，由于我将其更改为alertController，这部分代码也出现了错误：

var urlData: NSData? = NSURLConnection.sendSynchronousRequest(request, returningResponse:&response, error:reponseError)

    if ( urlData != nil ) {
        let res = response as! NSHTTPURLResponse!;

        NSLog("Response code: %ld", res.statusCode);

        if (res.statusCode >= 200 && res.statusCode < 300)
        {
            var responseData:NSString  = NSString(data:urlData!, encoding:NSUTF8StringEncoding)!

            NSLog("Response ==> %@", responseData);

            var error: NSError?

            let jsonData:NSDictionary = NSJSONSerialization.JSONObjectWithData(urlData!, options:NSJSONReadingOptions.MutableContainers , error: &error) as! NSDictionary

声明“调用中的额外参数‘错误’”，但是如果我删除了错误，那么它只会给我一个不同的错误，希望我删除)并添加一个，连续！

谢谢!

Answer 1

UIAlertView是UIAlertController的类是完全不同的，是而不是可互换的

UIAlertController的等效代码是

let alertController = UIAlertController(title: "Sign Up Failed!", 
                                      message: "Please enter Username and Password", 
                               preferredStyle: .Alert)
let okAction = UIAlertAction(title: "OK", style: .Default, handler: nil)
alertController.addAction(okAction)
self.presentViewController(alertController, animated: true, completion: nil)

要解决extra argument error，需要使用Swift 2的新错误处理语法

do {
   let jsonData = try NSJSONSerialization.JSONObjectWithData(urlData!, options:.MutableContainers) as! [String:AnyObject]
} catch {
    print(error)
}

如果不能保证jsonData成为字典(最好使用Swift本机类型)，则需要额外一行来检查带有可选绑定的类型

do {
   let jsonData = try NSJSONSerialization.JSONObjectWithData(urlData!, options:.MutableContainers)
   if jsonDict = jsonData as? [String:AnyObject] {
      // do something with the dictionary
   }
} catch {
    print(error)
}

PS：username.isEqualToString("")的一个更简单的语法是username.isEmpty

Answer 2

您必须将其呈现为UIViewController：

self.presentViewController(alertView, animated: true, completion: nil)

Answer 3

如果您创建一个没有任何操作按钮的警报框，请尝试以下代码。

let alertController = UIAlertController(title: "Alert", message: "This is an alert.", preferredStyle: .alert)


self.present(alertController, animated: true, completion: nil)

如果创建带有操作按钮的警报框，请尝试以下代码。

let alertController = UIAlertController(title: "Alert", message: "This is an alert.", preferredStyle: .alert)

let action1 = UIAlertAction(title: "Default", style: .default) { (action:UIAlertAction) in
    print("You've pressed default");
}

let action2 = UIAlertAction(title: "Cancel", style: .cancel) { (action:UIAlertAction) in
    print("You've pressed cancel");
}

let action3 = UIAlertAction(title: "Destructive", style: .destructive) { (action:UIAlertAction) in
    print("You've pressed the destructive");
}

alertController.addAction(action1)
alertController.addAction(action2)
alertController.addAction(action3)
self.present(alertController, animated: true, completion: nil)