Swift 有没有办法将KVP字典转换为字符串？

我正在尝试将从FireStore获取的

[String:Bool]

变量转换为字符串

例如：

var action = ["Nourishing":true, "Radiance":true]

致：

“滋养，光辉”

---

这可能吗？

如果您只想将所有键组合成一个字符串，则可以执行以下操作：

var action = ["Nourishing":true, "Radiance":true, "Whatever":false]
let keysAll = action.keys.joined(separator: ", ")
print(keysAll)

var action = ["Nourishing":true, "Radiance":true, "Whatever":false]
let keysTrue = action.filter { $0.value }.keys.joined(separator: ", ")
print(keysTrue)

var action = ["Nourishing":true, "Radiance":true, "Whatever":false]
let keysTrue = action.flatMap { $0.value ? $0.key : nil }.joined(separator: ", ")
print(keysTrue)

结果:

滋养，光辉，等等

如果只需要某些键，则首先需要根据需要过滤键/值。例如，如果只需要值为

true

的键，则可以执行以下操作：

var action = ["Nourishing":true, "Radiance":true, "Whatever":false]
let keysAll = action.keys.joined(separator: ", ")
print(keysAll)

var action = ["Nourishing":true, "Radiance":true, "Whatever":false]
let keysTrue = action.filter { $0.value }.keys.joined(separator: ", ")
print(keysTrue)

var action = ["Nourishing":true, "Radiance":true, "Whatever":false]
let keysTrue = action.flatMap { $0.value ? $0.key : nil }.joined(separator: ", ")
print(keysTrue)

结果:

滋养的、光辉的

或者，您也可以这样做：

var action = ["Nourishing":true, "Radiance":true, "Whatever":false]
let keysAll = action.keys.joined(separator: ", ")
print(keysAll)

var action = ["Nourishing":true, "Radiance":true, "Whatever":false]
let keysTrue = action.filter { $0.value }.keys.joined(separator: ", ")
print(keysTrue)

var action = ["Nourishing":true, "Radiance":true, "Whatever":false]
let keysTrue = action.flatMap { $0.value ? $0.key : nil }.joined(separator: ", ")
print(keysTrue)

---

使用reduce还有更难的解决方案

 let action = ["Nourishing":true, "Radiance":true] 

     let  data =   action.reduce("") { (result, dic) -> String in
            return result.count == 0  ? (result + "\(dic.key)") :  (result + ",\(dic.key)")
      }
     print(data)

---

你没有数组，你有字典。是的，这是可能的。您想要所有键还是只想要值为

true

的键？谢谢！！这正是我想要的！：D