Question

所以不确定这里有什么问题，但是在这种情况下swift引发了我的错误，并且没有给我foo[0]的输出

let bar: [String: Any] = [
    "questions":[
        ["id":1,"text":"What is your name?"],
        ["id":2,"text":"What is the last name?"]
    ],
    "status":"baz",
    "registered": true
]

let foo: [[String: Any]] = bar["questions"] as [[String: Any]]

foo[0]

虽然如果我这样做会有效 -

let bar: [String: String] = [
    "questions":[
        ["id":"1","text":"What is your name?"],
        ["id":"2","text":"What is the last name?"]
    ],
    "status":"AUTHENTICATE",
    "registered": true
]

let foo: [[String: String]] = bar["questions"] as [[String: String]]

foo[0]

我将我的ID更改为字符串（注意我没有触及布尔值）以及Any更改为字符串。

请为这种行为说清楚。

谢谢

Answer 1

如果你编写一个复杂的字典文字，如["id":1,"text":"What is your name?"]而没有以某种方式注释该类型，它将默认创建一个NSDictionary。

Swift中的

NSDictionary桥接到[NSObject: AnyObject]。强制将其强制转换为[String: Any]会因错误而崩溃＆＃34;无法重新解释不同大小的广告值＆＃34;。但是将NSDictionary投射到[String: AnyObject]会很好。

因此，您可以使用[String: AnyObject]或NSDictionary代替[String: Any]，也可以提示所需词典的类型，例如：

let bar: [String: Any] = [
    "questions":[
        ["id":1,"text":"What is your name?"] as [String: Any],
        ["id":2,"text":"What is the last name?"]
    ],
    "status":"baz",
    "registered": true
]

或者像这样

let bar: [String: Any] = [
    "questions":[
        ["id":1,"text":"What is your name?"],
        ["id":2,"text":"What is the last name?"]
    ] as [[String: Any]],
    "status":"baz",
    "registered": true
]

斯威夫特：＆＃34;任何＆＃34;没有成员命名元素

1 个答案: