我有一个由AnyObject组成的数组。我想遍历它,找到所有数组实例的元素。
我怎么能检查如果一个对象是一个给定的类型在Swift?
我有一个由AnyObject组成的数组。我想遍历它,找到所有数组实例的元素。
我怎么能检查如果一个对象是一个给定的类型在Swift?
当前回答
let originalArray : [Any?] = ["Hello", "World", 111, 2, nil, 3.34]
let strings = originalArray.compactMap({ $0 as? String })
print(strings)
//printed: ["Hello", "World"]
其他回答
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")
}
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")
}
}
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:
...
}
斯威夫特3:
class Shape {}
class Circle : Shape {}
class Rectangle : Shape {}
if aShape.isKind(of: Circle.self) {
}
如果你想检查一个特定的类型,你可以做以下事情:
if let stringArray = obj as? [String] {
// obj is a string array. Do something with stringArray
}
else {
// obj is not a string array
}
你可以使用"as!",如果obj不是[String]类型,就会抛出一个运行时错误。
let stringArray = obj as! [String]
你也可以一次检查一个元素:
let items : [Any] = ["Hello", "World"]
for obj in items {
if let str = obj as? String {
// obj is a String. Do something with str
}
else {
// obj is not a String
}
}