json - 如何在 Swift 中将 JSON 响应转换为字典格式

Question

如何转换此 JSON 响应，并在消息中包含数百个键，以便能够将其分配给变量类型 [AnyHashable: Any]

JSON

{
    "content": {
       "message": {
            "greet": "Hello world",
            "error": "There is an error",
            "thanks": "Thank you",
            ...
        }
    }
}

尝试

解析 JSON 后，我无法将 jsonData 分配给接受 [AnyHashable: Any] 的变量。我尝试将 message 类型更改为 [AnyHashable: Any]，但它不符合 Decodable。

// Model.swift

struct TestCase: Decodable {
    let content: Content
}

struct Content: Decodable {
    let message: [String: String]
}


// ViewController.swift

private var testVariable: [AnyHashable: Any]

private func loadJson(filename: String) {
    if let url = Bundle.main.url(forResource: filename, withExtension: "json") {
        do {
            let data = try Data(contentsOf: url)
            let decoder = JSONDecoder()
            let jsonData = try decoder.decode(TestCase.self, from: data)

            // Problem faced
            testVariable = jsonData // error

        } catch let jsonError {
            print("JsonError: ", jsonError)
        }
    }
}

Answer 1

你不能两者兼得。

您想用 Decodable 解析 JSON那么你必须声明 testVariable作为

private var testVariable : TestCase?

或者您想保留testVariable如[AnyHashable: Any]然后删除结构并使用传统的 JSONSerialization 解码 JSON

<罢工>

<罢工>

//print(s)
struct TestCase: Decodable {
    let content: Content
}

struct Content: Decodable {
    let message: [String: String]
}

<罢工>

// ViewController.swift

private var testVariable = [AnyHashable:Any]()

private func loadJson(filename: String) {
    if let url = Bundle.main.url(forResource: filename, withExtension: "json") {
        do {
            let data = try Data(contentsOf: url)
            testVariable = try JSONSerialization.jsonObject(with: data) as? [AnyHashable: Any] ?? [:]

        } catch {
            print("JsonError: ", error)
        }
    }
}