Swift 4 解码简单的根级 json 值

Question

根据JSON标准RFC 7159 ，这是有效的 json:

22

如何使用 swift4 的解码器将其解码为 Int？这不起作用

let twentyTwo = try? JSONDecoder().decode(Int.self, from: "22".data(using: .utf8)!)

Answer 1

它适用于很好的 JSONSerialization 和 .allowFragments阅读选项。来自documentation :

allowFragments

Specifies that the parser should allow top-level objects that are not an instance of NSArray or NSDictionary.

例子:

let json = "22".data(using: .utf8)!

if let value = (try? JSONSerialization.jsonObject(with: json, options: .allowFragments)) as? Int {
    print(value) // 22
}

然而，JSONDecoder 没有这样的选项并且不接受顶级不是数组或字典的对象。人们可以在 source code decode() 方法调用JSONSerialization.jsonObject() 没有任何选项:

open func decode<T : Decodable>(_ type: T.Type, from data: Data) throws -> T {
    let topLevel: Any
    do {
       topLevel = try JSONSerialization.jsonObject(with: data)
    } catch {
        throw DecodingError.dataCorrupted(DecodingError.Context(codingPath: [], debugDescription: "The given data was not valid JSON.", underlyingError: error))
    }

    // ...

    return value
}