xcode 使用 Swift 创建字典作为可选属性

Create a Dictionary as a optional property using Swift

提问by Sebastian

I created a swift class to test Dictionaries. So, I wrote the code below:

我创建了一个 swift 类来测试字典。所以，我写了下面的代码：

import Foundation

   class MyClass {

    var myFirstDictionary:[String :String]
    var myThirdDictionary:[String :String]?

    init(){
        var mySecondDictionary:[String :String] = [String :String]()
        mySecondDictionary["animal"] = "Monkey"
        mySecondDictionary.updateValue("something", forKey: "SomeKey")

        self.myFirstDictionary = [String :String]()
        addOneThingToSecondDictionary()
        addAnotherThingToSecondDictionary()

        self.myThirdDictionary! = [String :String]()
        addOneThingToThirdDictionary()
        addAnotherThingToThirdDictionary()
    }

    func addOneThingToSecondDictionary(){
        self.myFirstDictionary["animal"] = "Monkey"
    }

    func addAnotherThingToSecondDictionary(){
        self.myFirstDictionary.updateValue("Superman", forKey: "hero")
    }

    func addOneThingToThirdDictionary(){
        self.myThirdDictionary["animal"]! = "Monkey"
    }

    func addAnotherThingToThirdDictionary(){
        self.myThirdDictionary!.updateValue("Superman", forKey: "hero")
    }

  }

So, I got 3 errors referring to "myThirdDictionary":

所以，我收到了 3 个关于“myThirdDictionary”的错误：

• In the Dictionary initialization compiler said: Could not find an overload for 'init' that accepts the supplied arguments
• When I tried to add a Key/value pair in addOneThingToThirdDictionary(): '[String : String]?' does not have a member named 'subscript'
• When I tried to add a Key/value pair in addAnotherThingToThirdDictionary(): Immutable value of type '[String : String]' only has mutating members named 'updateValue'

• 在字典初始化编译器中说：找不到接受提供的参数的“init”的重载
• 当我尝试在addOneThingToThirdDictionary() 中添加键/值对时：'[String : String]?' 没有名为“下标”的成员
• 当我尝试在addAnotherThingToThirdDictionary() 中添加键/值对时： “[String : String]”类型的不可变值只有名为“updateValue”的变异成员

Any thoughts ?

有什么想法吗 ？

采纳答案by rickster

Some of these issues are conceptual errors, and some of them have to do with behaviors that changed in today's Xcode 6 beta 5 release. Running through them all:

其中一些问题是概念性错误，其中一些与今天的 Xcode 6 beta 5 版本中更改的行为有关。贯穿它们：

• This line compiles, but has a superfluous !:

  self.myThirdDictionary! = [String :String]()
  

  You don't need to unwrap an optional to assign to it -- it doesn't matter if its current contents are nilif you're providing new contents. Instead, just assign:

  self.myThirdDictionary = [String :String]()
  

• Similarly, this line fails because you're subscripting before unwrapping:

  self.myThirdDictionary["animal"]! = "Monkey"
  

  This is a problem because you could be subscripting nilif myThirdDictionaryhas not been initialized. Instead, subscript after checking/unwrapping the optional. As of beta 5, you can use mutating operators or methods through an optional check/unwrap, so the shortest and safest way to do this is:

  self.myThirdDictionary?["animal"] = "Monkey"
  

  If myThirdDictionaryis nil, this line has no effect. If myThirdDictionaryhas been initialized, the subscript-set operation succeeds.

• This line failed on previous betas because force-unwrapping produced an immutable value:

  self.myThirdDictionary!.updateValue("Superman", forKey: "hero")
  

  Now, it works -- sort of -- because you can mutate the result of a force-unwrap. However, force unwrapping will crash if the optional is nil. Instead, it's better to use the optional-chaining operator (which, again, you can now mutate through):

  self.myThirdDictionary?.updateValue("Superman", forKey: "hero")
  

• 这一行可以编译，但有一个多余的!：

  self.myThirdDictionary! = [String :String]()
  

  你不需要解开一个可选项来分配给它——nil如果你提供新的内容，它的当前内容是否无关紧要。相反，只需分配：

  self.myThirdDictionary = [String :String]()
  

• 同样，此行失败，因为您在解包之前添加了下标：

  self.myThirdDictionary["animal"]! = "Monkey"
  

  这是一个问题，因为nil如果myThirdDictionary尚未初始化，您可能正在下标。相反，在检查/解开可选项后下标。从 beta 5 开始，您可以通过可选的检查/解包来使用变异运算符或方法，因此执行此操作的最短和最安全的方法是：

  self.myThirdDictionary?["animal"] = "Monkey"
  

  如果myThirdDictionary是nil，则此行无效。如果myThirdDictionary已经初始化，则下标集操作成功。

• 这一行在之前的测试版上失败了，因为强制展开产生了一个不可变的值：

  self.myThirdDictionary!.updateValue("Superman", forKey: "hero")
  

  现在，它起作用了——有点——因为你可以改变强制解包的结果。但是，如果可选项为 ，则强制展开将崩溃nil。相反，最好使用可选链运算符（同样，您现在可以通过它进行变异）：

  self.myThirdDictionary?.updateValue("Superman", forKey: "hero")
  

Finally, you have a lot of things in this code that can be slimmed down due to type and scope inference. Here it is with all the issues fixed and superfluous bits removed:

最后，由于类型和范围推断，您可以在此代码中精简很多内容。这是所有已修复的问题并删除了多余的部分：

class MyClass {

    var myFirstDictionary: [String: String]
    var myThirdDictionary: [String: String]?

    init(){
        var mySecondDictionary: [String: String] = [:]
        mySecondDictionary["animal"] = "Monkey"
        mySecondDictionary.updateValue("something", forKey: "SomeKey")

        myFirstDictionary = [:]
        addOneThingToSecondDictionary()
        addAnotherThingToSecondDictionary()

        // uncomment to see what happens when nil
        myThirdDictionary = [:]
        addOneThingToThirdDictionary()
        addAnotherThingToThirdDictionary()
    }

    func addOneThingToSecondDictionary(){
        myFirstDictionary["animal"] = "Monkey"
    }

    func addAnotherThingToSecondDictionary(){
        myFirstDictionary.updateValue("Superman", forKey: "hero")
    }

    func addOneThingToThirdDictionary(){
        myThirdDictionary?["animal"] = "Monkey"
    }

    func addAnotherThingToThirdDictionary(){
        myThirdDictionary?.updateValue("Superman", forKey: "hero")
    }

}

(Changes: Foundationimport unused, empty dictionary literal instead of repeated type info)

（更改：Foundation导入未使用的空字典文字而不是重复的类型信息）

回答by Zain Shahzad

Simple but effective

简单但有效

let like = (dic.value(forKey: "likes") as? NSDictionary)
print(like?.count)

Or if you want get complete Dictionary count then use:

或者，如果您想获得完整的字典计数，请使用：

 if let mainDic = dic as? NSDictionary{ 
    print(mainDic.count) 
    //add your code
}