如果你有这样的响应:

{
  "registeration_method": "email",
  "is_stucked": true,
  "individual": {
    "id": 24099,
    "first_name": "ahmad",
    "last_name": "zozoz",
    "email": null,
    "mobile_number": null,
    "confirmed": false,
    "avatar": "http://abc-abc-xyz.amazonaws.com/images/placeholder-profile.png",
    "doctor_request_status": 0
  },
  "max_number_of_confirmation_trials": 4,
  "max_number_of_invalid_confirmation_trials": 12
}

你想检查值is_stuck将被读取为AnyObject，你所要做的就是这个

if let isStucked = response["is_stucked"] as? Bool{
  if isStucked{
      print("is Stucked")
  }
  else{
      print("Not Stucked")
 }
}

只是为了完整起见，基于公认的答案和其他一些答案:

let items : [Any] = ["Hello", "World", 1]

for obj in items where obj is String {
   // obj is a String. Do something with str
}

但是你也可以(compactMap也“映射”过滤器没有的值):

items.compactMap { $0 as? String }.forEach{ /* do something with $0 */ ) }

以及一个使用switch的版本:

for obj in items {
    switch (obj) {
        case is Int:
           // it's an integer
        case let stringObj as String:
           // you can do something with stringObj which is a String
        default:
           print("\(type(of: obj))") // get the type
    }
}

但回到问题上来，检查它是否是一个数组(即[String]):

let items : [Any] = ["Hello", "World", 1, ["Hello", "World", "of", "Arrays"]]

for obj in items {
  if let stringArray = obj as? [String] {
    print("\(stringArray)")
  }
}

或者更一般地说(见另一个问题的答案):

for obj in items {
  if obj is [Any] {
    print("is [Any]")
  }

  if obj is [AnyObject] {
    print("is [AnyObject]")
  }

  if obj is NSArray {
    print("is NSArray")
  }
}

我有两种方法:

if let thisShape = aShape as? Square 

Or:

aShape.isKindOfClass(Square)

下面是一个详细的例子:

class Shape { }
class Square: Shape { } 
class Circle: Shape { }

var aShape = Shape()
aShape = Square()

if let thisShape = aShape as? Square {
    println("Its a square")
} else {
    println("Its not a square")
}

if aShape.isKindOfClass(Square) {
    println("Its a square")
} else {
    println("Its not a square")
}

编辑:3现在:

let myShape = Shape()
if myShape is Shape {
    print("yes it is")
}

myObject一样吗?如果myObject不是String, String返回nil。否则，它返回一个字符串?，所以你可以使用myObject访问字符串本身!，或者使用myObject强制转换它!作为字符串安全。

你可以使用这个函数，然后调用它:

func printInfo(_ value: Any) {
    let t = type(of: value)
    print("'\(value)' of type '\(t)'")
}

例如:printInfo(data)

数据类型为“125字节”

在Swift 2.2 - 5你现在可以做:

if object is String
{
}

然后过滤你的数组:

let filteredArray = originalArray.filter({ $0 is Array })

如果你有多个类型需要检查:

    switch object
    {
    case is String:
        ...

    case is OtherClass:
        ...

    default:
        ...
    }