我根据@John R Perry 的解决方案制作了一个递归函数,该函数更深入地研究了对象的属性。它还需要一个参数来限制它的深度(默认为Int.max)以帮助防止stackoverflow:
protocol Loopable {
func allProperties(limit: Int) [String: Any]
}
extension Loopable {
func allProperties(limit: Int = Int.max) [String: Any] {
return props(obj: self, count: 0, limit: limit)
}
private func props(obj: Any, count: Int, limit: Int) -> [String: Any] {
let mirror = Mirror(reflecting: obj)
var result: [String: Any] = [:]
for (prop, val) in mirror.children {
guard let prop = prop else { continue }
if limit == count {
result[prop] = val
} else {
let subResult = props(obj: val, count: count + 1, limit: limit)
result[prop] = subResult.count == 0 ? val : subResult
}
}
return result
}
}
我不再检查对象是否为class 或struct,因为参数不是class 或struct 是递归函数的基本情况,而且更容易手动处理而不是出错。
测试它:
class C {
var w = 14
}
class B: Loopable {
var x = 12
var y = "BHello"
var z = C()
static func test() -> String {
return "Test"
}
}
class A: Loopable {
var a = 1
var c = 10.0
var d = "AHello"
var e = true
var f = B()
var g = [1,2,3,4]
var h: [String: Any] = ["A": 0, "B": "Dictionary"]
var i: Int?
}
print(A().allProperties())
打印:
["e": true, "g": [1, 2, 3, 4], "f": ["z": ["w": 14], "x": 12, "y": "BHello"], "h": ["A": 0, "B": "Dictionary"], "c": 10.0, "i": nil, "d": "AHello", "a": 1]
(字典是无序的,所以如果你得到不同的顺序,那就是为什么)