Swift 4 decode simple root level json value

Question

According to the JSON standard RFC 7159, this is valid json:

22

How do I decode this into an Int using swift4's decodable? This does not work

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

Answer 1

It works with good ol' JSONSerialization and the .allowFragments reading option. From the documentation:

allowFragments

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

Example:

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

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

However, JSONDecoder has no such option and does not accept top-level objects which are not arrays or dictionaries. One can see in the source code that the decode() method calls JSONSerialization.jsonObject() without any option:

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
}

Answer 2

In iOS 13.1+ and macOS 10.15.1+ JSONDecoder can handle primitive types on root level.

See the latest comments (Oct 2019) in the linked article underneath Martin's answer.