Encode nil value as null with JSONEncoder

Question

I'm using Swift 4's JSONEncoder. I have a Codable struct with an optional property, and I'd like this property to show up as null value in the produced JSON data when the value is nil. However, JSONEncoder discards the property and does not add it to the JSON output. Is there a way to configure JSONEncoder so that it preserves the key and sets it to null in this case?

Example

The code snippet below produces {"number":1}, but I'd rather like it to give me {"string":null,"number":1}:

struct Foo: Codable {
  var string: String? = nil
  var number: Int = 1
}

let encoder = JSONEncoder()
let data = try! encoder.encode(Foo())
print(String(data: data, encoding: .utf8)!)

Answer 1

I ran into the same problem. Solved it by creating a dictionary from the struct without using JSONEncoder. You can do this in a relatively universal way. Here's my code:

struct MyStruct: Codable {
    let id: String
    let regionsID: Int?
    let created: Int
    let modified: Int
    let removed: Int?


    enum CodingKeys: String, CodingKey, CaseIterable {
        case id = "id"
        case regionsID = "regions_id"
        case created = "created"
        case modified = "modified"
        case removed = "removed"
    }

    var jsonDictionary: [String : Any] {
        let mirror = Mirror(reflecting: self)
        var dic = [String: Any]()
        var counter = 0
        for (name, value) in mirror.children {
            let key = CodingKeys.allCases[counter]
            dic[key.stringValue] = value
            counter += 1
        }
        return dic
    }
}

extension Array where Element == MyStruct {
    func jsonArray() -> [[String: Any]] {
        var array = [[String:Any]]()
        for element in self {
            array.append(element.jsonDictionary)
        }
        return array
    }
}

You can do this without the CodingKeys (if the table attribute names on server side are equal to your struct property names). In that case just use the 'name' from mirror.children.

If you need CodingKeys don't forget to add the CaseIterable protocol. That makes it possible to use the allCases variable.

Be careful with nested structs: E.g. if you have a property with a custom struct as type, you need to convert that to a dictionary too. You can do this in the for loop.

The Array extension is required if you want to create an array of MyStruct dictionaries.

Answer 2

Yes, but you'll have to write your own encoder; you can't use the default one.

struct Foo: Codable {
    var string: String? = nil
    var number: Int = 1

    func encode(to encoder: Encoder) throws {
        var container = encoder.container(keyedBy: CodingKeys.self)
        try container.encode(number, forKey: .number)
        try container.encode(string, forKey: .string)
    }
}

Encoding an optional directly will encode a null, like you're looking for.

If this is an important use case for you, you may consider opening a defect at bugs.swift.org to ask for a new OptionalEncodingStrategy flag to be added on JSONEncoder to match the existing DateEncodingStrategy, etc. (See below why this is likely impossible to actually implement in Swift today, but getting into the tracking system is still useful as Swift evolves.)

---

Edit: To Paulo's questions below, this dispatches to the generic encode<T: Encodable> version because Optional conforms to Encodable. This is implemented in Codable.swift this way:

extension Optional : Encodable /* where Wrapped : Encodable */ {
    @_inlineable // FIXME(sil-serialize-all)
    public func encode(to encoder: Encoder) throws {
        assertTypeIsEncodable(Wrapped.self, in: type(of: self))

        var container = encoder.singleValueContainer()
        switch self {
        case .none: try container.encodeNil()
        case .some(let wrapped): try (wrapped as! Encodable).__encode(to: &container)
        }
    }
}

This wraps the call to encodeNil, and I think letting stdlib handle Optionals as just another Encodable is better than treating them as a special case in our own encoder and calling encodeNil ourselves.

Another obvious question is why it works this way in the first place. Since Optional is Encodable, and the generated Encodable conformance encodes all the properties, why does "encode all the properties by hand" work differently? The answer is that the conformance generator includes a special case for Optionals:

// Now need to generate `try container.encode(x, forKey: .x)` for all
// existing properties. Optional properties get `encodeIfPresent`.
...

if (varType->getAnyNominal() == C.getOptionalDecl() ||
    varType->getAnyNominal() == C.getImplicitlyUnwrappedOptionalDecl()) {
  methodName = C.Id_encodeIfPresent;
}

This means that changing this behavior would require changing the auto-generated conformance, not JSONEncoder (which also means it's probably really hard to make configurable in today's Swift....)