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Question

我想从配置文件字典中获取地址，但出现错误 "type any? has no subscript members"

var address:[[String : Any]] = [["Address": "someLocation", "City": "ABC","Zip" : 123],["Address": "someLocation", "City": "DEF","Zip" : 456]]
var profile:[String : Any] = ["Name": "Mir", "Age": 10, "Addresses": address]
profile["Addresses"][0]     <-----------------type any? has no subscript members

如何修复它并获取地址？非常感谢。

Answer 1

当您使用 "Addresses" 订阅配置文件时，您将获得一个 Any 实例。您选择使用 Any 来适应同一数组中的各种类型，这导致了类型擦除的发生。您需要将结果转换回其真实类型 [[String: Any]]，以便它知道 Any 实例表示 Array。然后你就可以下标了：

func f() {
    let address: [[String : Any]] = [["Address": "someLocation", "City": "ABC","Zip" : 123],["Address": "someLocation", "City": "DEF","Zip" : 456]]
    let profile: [String : Any] = ["Name": "Mir", "Age": 10, "Addresses": address]

    guard let addresses = profile["Addresses"] as? [[String: Any]] else {
        // Either profile["Addresses"] is nil, or it's not a [[String: Any]]
        // Handle error here
        return
    }

    print(addresses[0])
}

虽然这很笨拙，而且首先使用字典并不是一个非常合适的情况。

在这种情况下，您的字典具有一组固定的键，结构是更合适的选择。它们是强类型的，因此您不必从 Any 进行向上和向下转换，它们具有更好的性能，并且更易于使用。试试这个：

struct Address {
    let address: String
    let city: String
    let zip: Int
}

struct Profile {
    let name: String
    let age: Int
    let addresses: [Address]
}

let addresses = [
    Address(
        address: "someLocation"
        city: "ABC"
        zip: 123
    ),
    Address(
        address: "someLocation"
        city: "DEF"
        zip: 456
    ),
]

let profile = Profile(name: "Mir", age: 10, addresses: addresses)

print(profile.addresses[0]) //much cleaner/easier!

Answer 2

您应该重新考虑您选择构建 adress 和 profile 的方式；参见例如.

对于技术讨论，您可以提取您知道包含 [String: Any] 字典的 profile 的 Any 成员Any数组；通过顺序尝试将 profile["Addresses"] 类型转换为 [Any]，然后逐个元素（尝试）转换为 [String: Any]:

if let adressDictsWrapped = profile["Addresses"] as? [Any] {
    let adressDicts = adressDictsWrapped.flatMap{ [=10=] as? [String: Any] }
    print(adressDicts[0]) // ["Zip": 123, "City": "ABC", "Address": "someLocation"]
    print(adressDicts[1]) // ["Zip": 456, "City": "DEF", "Address": "someLocation"]
}

或者，没有中间步骤...

if let adressDicts = profile["Addresses"] as? [[String: Any]] {
   print(adressDicts[0]) // ["Zip": 123, "City": "ABC", "Address": "someLocation"]
   print(adressDicts[1]) // ["Zip": 456, "City": "DEF", "Address": "someLocation"]
}

但这只是尝试类型化转换的一个小教训（-> 不要这样做）。

Answer 3

如果您按照之前的建议重新考虑您的设计，我同意。为了便于讨论，您可以执行以下操作来实现您的目标。

var address:[[String : Any]] = [["Address": "someLocation", "City": "ABC","Zip" : 123],["Address": "someLocation", "City": "DEF","Zip" : 456]]
var profile:[String : Any] = ["Name": "Mir", "Age": 10, "Addresses": address]
if let allAddresses = profile["Addresses"] as? [[String:Any]] {
    print("This are all the address \(allAddresses[0])")
    }

键入任何？没有下标成员

type any? has no subscript members

arrays

swift

dictionary

any

anyobject