Swift 4 decode simple root level json value

Question

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

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)!)

this is not a valid json, a json is made by key and value pairs and it's just a value. — Luiz Fernando Salvaterra
– Luiz Fernando Salvaterra, Commented Oct 16, 2017 at 10:56
If you wrap your try in a Do..Catch and handle the error, it will tell you that your json format is not correct. That is not a valid JSON, why are you insisting on this? — Desdenova
– Desdenova, Commented Oct 16, 2017 at 11:19
JSONSerialization has a allowFragments reading option for that purpose, but the new JSONDecoder has no such option. — Martin R
– Martin R, Commented Oct 16, 2017 at 11:19
@Desdenova I am not insisting, it is valid according to the RFC 7159 standard which is definitely being used widely. If what you are saying is true I guess this means swift is not using it though. — mikkelam
– mikkelam, Commented Oct 16, 2017 at 11:22

Community · Accepted Answer · 2020-06-20 09:12:55Z

15

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
}

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered Oct 16, 2017 at 11:30

Martin R

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2 Comments

Martin R Over a year ago

You can try and file a bug report at Apple.

mikkelam Over a year ago

I forgot to write, this but I submitted an issue here bugs.swift.org/browse/SR-6163

vadian · Accepted Answer · 2019-12-09 22:42:34Z

5

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.

answered Dec 9, 2019 at 22:42

vadian

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