如果需要处理函数，则无法将 alertController 移动到 UIViewController 扩展

Question

在我的整个程序中，我多次调用 UIAlertControllers。为了让事情更容易阅读，如果 AlertController 只有一个 "OK" 动作且 nil 作为处理程序，我就能够成功地分解它。

如果我想传递一个我想 link 作为 UIAlertAction 的函数，我正在努力做同样的事情。 func showOKCancelAlertController 不编译

showOKAlertController(title: "Network Error", message: "Unable to download photos") //This is an example of how I call Alert Controller throughout my code.

extension UIViewController {
    func showOKAlertController(title: String, message: String){
        let myAlertController = UIAlertController(title: title, message: message, preferredStyle: .alert)
        myAlertController.addAction(UIAlertAction(title: "OK", style: .default, handler: nil))
        present(myAlertController, animated: true)
    }

    func showOKCancelAlertController(title: String, message: String, okFunction: UIAlertAction ){
        let myAlertController = UIAlertController(title: title, message: message, preferredStyle: .alert)
        myAlertController.addAction(UIAlertAction(title: "OK", style: .default, handler: okFunction))
        myAlertController.addAction(UIAlertAction(title: "Cancel", style: .default, handler: nil))
        present(myAlertController, animated: true)
    }
}

Answer 1

处理程序预期类型是 ((UIAlertAction) -> Swift.Void)? 而不是 UIAlertAction

将您的代码更新为

func showOKCancelAlertController(title: String, message: String, okFunction: ((UIAlertAction) -> Swift.Void)? ){
     let myAlertController = UIAlertController(title: title, message: message, preferredStyle: .alert)
     myAlertController.addAction(UIAlertAction(title: "OK", style: .default, handler: okFunction))
     myAlertController.addAction(UIAlertAction(title: "Cancel", style: .default, handler: nil))
     present(myAlertController, animated: true)
}

Answer 2

可能是

extension UIViewController { 
    func showOKAlertController(title: String, message: String,ok:@escaping(() -> ())) {
        let myAlertController = UIAlertController(title: title, message: message, preferredStyle: .alert)
        myAlertController.addAction(UIAlertAction(title: "ok", style: .default, handler: { (al) in
            ok()
        }))
        myAlertController.addAction(UIAlertAction(title: "Cancel", style: .default, handler: nil))
        present(myAlertController, animated: true)
    }
}

showOKAlertController(title: "This is title", message: "this is message") {
   print("ok")
}

Answer 3

显然，您面临的错误是：

Cannot convert value of type 'UIAlertAction' to expected argument type '((UIAlertAction) -> Void)?'

那是因为 okFunction 类型 UIAlertAction 的参数，这是不正确的部分。您应该让 okFunction 成为 ((UIAlertAction) -> Void)? 类型：

func showOKCancelAlertController(title: String, message: String, okFunction: ((UIAlertAction) -> Void)?) {
    let myAlertController = UIAlertController(title: title, message: message, preferredStyle: .alert)
    myAlertController.addAction(UIAlertAction(title: "OK", style: .default, handler: okFunction))
    myAlertController.addAction(UIAlertAction(title: "Cancel", style: .default, handler: nil))
    present(myAlertController, animated: true)
}

那是因为 UIAlertAction 初始化签名是：

init(title: String?, style: UIAlertAction.Style, handler: ((UIAlertAction) -> Void)? = nil),

handler 参数需要 ((UIAlertAction) -> Void)?。

因此，您称它为：

showOKCancelAlertController(title: "title", message: "message") { _ in
    print("here is what to do when tapping the OK button")
}

此外，如果“确定”按钮没有任何操作，您可以为 okFunction 参数提供默认 nil 值：

func showOKCancelAlertController(title: String, message: String, okFunction: ((UIAlertAction) -> Void)? = nil)

并将其命名为：

showOKCancelAlertController(title: "title", message: "message")

实际上，这对您的情况来说是一件很酷的事情：此时您甚至不需要实现两种不同的方法，您可以只实现一种方法并传递 okFunction 参数仅在需要时才使用它！示例：

func showAlertController(title: String, message: String, okFunction: ((UIAlertAction) -> Void)? = nil) {
    let myAlertController = UIAlertController(title: title, message: message, preferredStyle: .alert)
    myAlertController.addAction(UIAlertAction(title: "OK", style: .default, handler: okFunction))
    if let okFun = okFunction {
        myAlertController.addAction(UIAlertAction(title: "Cancel", style: .default, handler: okFun))
    }

    present(myAlertController, animated: true)
}

如果要一键显示：

showAlertController(title: "title", message: "message")

以及两个按钮：

showAlertController(title: "title", message: "message") { _ in
    // ...
}

Answer 4

按照我的方式改进你的代码

这可以灵活处理完成处理程序

extension UIViewController {
    func showOKAlertController(title: String, message: String? = nil, okCompletion: @escaping (() -> ()) = {}, presentCompletion: @escaping (() -> ()) = {}) {
        let myAlertController = UIAlertController(title: title, message: message, preferredStyle: .alert)
        let okAction = UIAlertAction(title: "OK", style: .default) { (action: UIAlertAction) in
            okCompletion()
        }
        myAlertController.addAction(okAction)
        DispatchQueue.main.async {
            self.present(alertController, animated: true) {
                presentCompletion()
            }
        }
    }

    func showOKCancelAlertController(title: String, message: String? = nil, okCompletion: @escaping (() -> ()) = {}, cancelCompletion: @escaping (() -> ()) = {}, presentCompletion: @escaping (() -> ()) = {}) {
        let myAlertController = UIAlertController(title: title, message: message, preferredStyle: .alert)
        let okAction = UIAlertAction(title: "OK", style: .default) { (action: UIAlertAction) in
            okCompletion()
        }
        let cancelAction = UIAlertAction(title: "Cancel", style: .cancel) { (cancel: UIAlertAction) in
            cancelCompletion()
        }
        myAlertController.addAction(okAction)
        myAlertController.addAction(cancelAction)
        DispatchQueue.main.async {
            self.present(alertController, animated: true) {
                presentCompletion()
            }
        }
    }
}

showOKAlertController(title: "message")

或

showOKAlertController(title: "message", message: "message")

或

showOKCancelAlertController(title: "message", message: nil, okCompletion: {
    // ok completion handling
}) {
    // present completion handling
}

或者你可以从我的 gist

中参考上面有用的内容

如果需要处理函数，则无法将 alertController 移动到 UIViewController 扩展

Unable to move alertController to UIViewController extension, if it requires handler function

ios

swift

uialertcontroller