Swift 4：递归地在嵌套的动态Dictionary <String，Any>中查找值

Question

在给定的Dictionary中，我需要找到[String : Any]给定键的嵌套Dictionary（）。

字典的一般结构（例如嵌套级别，值类型）是未知的，并且是动态给出的。[1]

在此子词典中，有一个需要获取的键“值”（不询问）的给定值。

这是一个例子：

let theDictionary: [String : Any] =
  [ "rootKey" :
    [ "child1Key" : "child1Value",
      "child2Key" : "child2Value",
      "child3Key" :
        [ "child3SubChild1Key" : "child3SubChild1Value",
          "child3SubChild2Key" :
              [ "comment" : "child3SubChild2Comment", 
                 "value" : "child3SubChild2Value" ]
        ],
      "child4Key" :
        [ "child4SubChild1Key" : "child4SubChild1Value",
          "child4SubChild2Key" : "child4SubChild2Value",
          "child4SubChild3Key" :
            [ "child4SubChild3SubChild1Key" :
                [ "value" : "child4SubChild3SubChild1Value", 
                  "comment" : "child4SubChild3SubChild1Comment" ]
            ]
        ]
    ]
  ]

通过蛮力和伪记忆，我设法一起破解了一个功能，该功能遍历整个Dictionary并获取给定键的值：

func dictionaryFind(_ needle: String, searchDictionary: Dictionary<String, Any>) -> String? {

  var theNeedleDictionary = Dictionary<String, Any>()

    func recurseDictionary(_ needle: String, theDictionary: Dictionary<String, Any>) -> Dictionary<String, Any> {
      var returnValue = Dictionary<String, Any>()
      for (key, value) in theDictionary {
        if value is Dictionary<String, Any> {
          if key == needle {
            returnValue = value as! Dictionary<String, Any>
            theNeedleDictionary = returnValue
            break
          } else {
              returnValue =  recurseDictionary(needle, theDictionary: value as! Dictionary<String, Any>)
            }
        }
     }
     return returnValue
    }
  // Result not used
  _ = recurseDictionary(needle, theDictionary: searchDictionary)

  if let value = theNeedleDictionary["value"] as? String {
    return value
  }
  return nil
}

到目前为止，该方法有效。（为您的游乐场测试带来乐趣：

let theResult1 = dictionaryFind("child3SubChild2Key", searchDictionary: theDictionary)
print("And the result for child3SubChild2Key is: \(String(describing: theResult1!))")

let theResult2 = dictionaryFind("child4SubChild3SubChild1Key", searchDictionary: theDictionary)
print("And the result for child4SubChild3SubChild1Key is: \(String(describing: theResult2!))")

let theResult3 = dictionaryFind("child4Key", searchDictionary: theDictionary)
print("And the result for child4Key is: \(String(describing: theResult3))")

）。

我的问题在这里：

什么是更干净，简洁，“快捷”的方法来遍历字典，尤其是在找到所需的密钥后完全脱离常规？

甚至可以使用Dictionary扩展实现解决方案吗？

谢谢大家！

[1]如从字典中删除嵌套键中所述的KeyPath 是不可行的。

Answer 1

更为紧凑的递归解决方案可能是：

func search(key:String, in dict:[String:Any], completion:((Any) -> ())) {
    if let foundValue = dict[key] {
        completion(foundValue)
    } else {
        dict.values.enumerated().forEach {
            if let innerDict = $0.element as? [String:Any] {
                search(key: key, in: innerDict, completion: completion)
            }
        }
    }
}

用法是：

search(key: "child3SubChild2Key", in: theDictionary, completion: { print($0) }) 

这使：

["comment": "child3SubChild2Comment", "value": "child3SubChild2Subchild1Value"]

或者，如果您不想使用闭包，则可以使用以下代码：

extension Dictionary {
    func search(key:String, in dict:[String:Any] = [:]) -> Any? {
        guard var currDict = self as? [String : Any]  else { return nil }
        currDict = !dict.isEmpty ? dict : currDict

        if let foundValue = currDict[key] {
            return foundValue
        } else {
            for val in currDict.values {
                if let innerDict = val as? [String:Any], let result = search(key: key, in: innerDict) {
                    return result
                }
            }
            return nil
        }
    }
}

用法是：

let result = theDictionary.search(key: "child4SubChild3SubChild1Key")
print(result) // ["comment": "child4SubChild3SubChild1Comment", "value": "child4SubChild3SubChild1Value"]