引用方法的 Objective-C 选择器

Question

从Swift 2.2开始，以下代码给出了警告：

No method declared with Objective-C selector'sync'

if let tabBarController = segue.destinationViewController as? TabBarController {
  tabBarController.navigationItem.rightBarButtonItem = 
    UIBarButtonItem(title: "Upload", 
                    style: .Plain, 
                    target: tabBarController, 
                    action: "sync")

我应该用什么替换action: "sync"它来消除警告？

我试过：

Selector("sync") // The Xcode provided fix which yields the same warning
#selector(tabBarController.sync()) // Error: Argument of '#selector' does not refer to initializer or method 
Selector(tabBarController.sync()) // No error/warning but doesn't fire sync function

Answer 1

我认为您放错了操作函数 "sync"。将它保存在 TabBarController 中，因为您已将 TabBarController 的实例用作目标。下面的代码将起作用：

tabBarController.navigationItem.rightBarButtonItem = UIBarButtonItem(title: "Upload", style: .Plain, target: tabBarController, action: "sync:")

在 TabBarController 中保留以下功能：

func sync(sender: AnyObject){
    //your code here
}

希望它能解决您的问题。 :)

Answer 2

为了解决您的问题，请首先阅读 documentation 关于 Swift2.2 中的选择器的新内容。

示例：使用 #selector(CLASS.sync) 而不是 Selector("sync")。其中 CLASS 实际 class 包含此方法。

之所以这样做是因为：

The use of string literals for selector names is extremely error-prone: there is no checking that the string is even a well-formed selector, much less that it refers to any known method, or a method of the intended class. Moreover, with the effort to perform automatic renaming of Objective-C APIs, the link between Swift name and Objective-C selector is non-obvious. By providing explicit "create a selector" syntax based on the Swift name of a method, we eliminate the need for developers to reason about the actual Objective-C selectors being used.

引用方法的 Objective-C 选择器

Referencing the Objective-C selector of a method

swift

swift2.2