swift可选链接与演员 [英] swift optional chaining with cast

问题描述

下面的代码突出显示了我将可选链接和强制转换与Apple的快速语言相结合的问题

The code below highlights a problem I'm having combining optional chaining and casts with Apple's swift language

import Foundation
import CoreData

class MyExample {

    var detailItem: NSManagedObject?

    func example() {
        //In the actual implementation it is assigned to a UITableViewCell textLabel.text with the same result.
        let name: String = self.detailItem?.valueForKey("name") as String
    }
}

上述结果为：

'AnyObject'不能转换为'String'

'AnyObject' is not convertible to 'String'

我可以通过使类变量成为隐式解包的可选项来实现，如下所示：

I am able to do this by making the class variable an implicitly unwrapped optional as follows:

class MyExample2 {

    var detailItem: NSManagedObject!

    func example() {
        let name: String = self.detailItem.valueForKey("name") as String
    }
}

这样可行，但现在这个变量并不反映我的真实世界需要它是可选的

This works, but now this variable doesn't reflect my real world need for it to be optional

下一个示例也适用于detailItem作为可选项：

This next example also works with detailItem as optional:

class MyExample3 {

    var detailItem: NSManagedObject?

    func example() {
        let name: String = self.detailItem?.valueForKey("name").description as String
    }
}

以上问题是我必须使用说明。使用此功能的输出向用户显示通常不是一个好主意（根据Apple，只是常识）

The problem with the above is I had to use description. It's generally not a good idea to use output from this function to show to the user (according to Apple, and just common sense)

以上也只能起作用，因为我 am 正在寻找一个字符串。如果我需要一个对象，它将无法工作。

The above also only works because I am looking for a string. If I needed an object, it wouldn't work.

作为一个兴趣点，这个例子引发了一个不同的错误：

As a point of interest this example throws a different error:

class MyExample4 {

    var detailItem: NSManagedObject?

    func example() {
        let name: String = self.detailItem?.valueForKey("name")
    }
}

上述抛出：

不能find member'valueForKey'

Could not find member 'valueForKey'

NSmanagedObject显然有一个valueForKey。

NSmanagedObject clearly has a valueForKey.

尝试还有一件事，我发现了一个潜在的解决方案：

Trying one more thing, I discovered a potential solution:

class MyExamplePotentialSolution {

    var detailItem: NSManagedObject?

    func example() {
        let name: NSString = self.detailItem?.valueForKey("name") as NSString
    }
}

上面的问题是，当它实际分配给UITableViewCell detailTextLabel.text属性时它不起作用。

The problem with the above, is it doesn't work when actually assigned to a UITableViewCell detailTextLabel.text attribute.

任何想法？

更新答案

最简单的现实世界用法是：

The simplest real world usage turned out to be this:

cell.detailTextLabel.text = self.detailItem?.valueForKey("name") as? NSString

关键是AnyObject不能直接转换为本机swift类型。正如接受的答案所示，它可以从NSString转换为本机Swift字符串。在这种情况下，它没有必要。

The key is AnyObject can't be cast to a native swift type directly. As the accepted answer showed, it can be cast as a native Swift string from NSString. It just isn't necessary in this case.

推荐答案

有两个问题：第一个是在这一行：

There are 2 problems: the first is that in this line:

let name: String = self.detailItem?.valueForKey("name") as String

右边的部分是可选的（ detailItem 是可选的），所以无论表达式返回什么，它不能分配给类型字符串

the right part is an optional (detailItem is optional), so whatever the expression returns, it cannot be assigned to a non-optional variable having type String

的非可选变量第二个问题是 valueForKey 返回 AnyObject！，它可以是任何类的实例 - 但是 String 不是一个类，你需要任何才能将它转换为 String 。
我假设 NSManagedObject 返回 NSString ，因此您可以通过以下行获得所需：

The second problem is that valueForKey returns AnyObject!, which can be an instance of any class - but String is not a class, you'd need Any in order to be able to cast that into a String. I presume that NSManagedObject returns an NSString, so you can achieve what you need with this line:

let name: String? = (self.detailItem?.valueForKey("name") as? NSString) as? String

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