Convert null values to default strings in parsing JSON using JSONDecoder in Swift

Question

I am trying to parse some JSON in Swift using JSONDecoder where the JSON occasionally has null values. I would like to put in a default instead. The following allows me to handle it but the nulls cause problems later.

struct Book: Codable {
        let title : String
        let author: String?
    }

Is there a way to do something like (following does not compile), perhaps using an initializer?:

struct Book: Codable {
        let title : String
        let author: String ?? "unknown"
    }

Thanks for any suggestions

dktaylor · Accepted Answer · 2020-06-10 20:44:54Z

6

This could be address by manually decoding as described here.

The other way to go would be to have the stored properties reflect the data exactly, and then have a computed var for the case of providing a non-optional value.

struct Book: Codable {
    let title : String
    let author: String?

    var displayAuthor: String {
        return author ?? "unknown"
    }
}

The other reason this might be appealing is it preserves the optional value should you need to check if the value exists at all in the future.

answered Jun 10, 2020 at 20:44

dktaylor

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1 Comment

zztop Over a year ago

Like that it preserves optional

Frankenstein · Accepted Answer · 2020-06-10 20:59:56Z

1

You can achieve this using the custom init(decoder:) method definition. Use the decodeIfPresent API and give the property your desired default value if the try fails. Or you can use the computed property method mentioned by @dktaylor. Here the code you need:

struct Book {
    let title : String
    let author: String

    enum CodingKeys: String, CodingKey {
        case title, author
    }
}

extension Book: Codable {
    init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)
        title = try container.decode(String.self, forKey: .title)
        author = try container.decodeIfPresent(String.self, forKey: .author) ?? "unknown"
    }
}

answered Jun 10, 2020 at 20:59

Frankenstein

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Comments

Sander Saelmans · Accepted Answer · 2020-06-10 22:13:23Z

You can also achieve this with a property wrapper:

@propertyWrapper
struct OnNil<T> {
    let value: T
    init(_ value: T) {
        self.value = value
    }

    private var _wrappedValue: T?
    var wrappedValue: T! {
        get { _wrappedValue ?? value }
        set { _wrappedValue = newValue }
    }
}

struct SomeStruct {
    let title : String

    @OnNil("unknown") 
    let author: String!

}

The benefits of using a property wrapper like this is that you don't have to soil your object with utility methods, and you don't have to fiddle with the decode function. The downside obviously is that the syntax looks kind of odd, you have to make some variables implicitly unwrapped optional, and it makes your code slightly harder to debug due to the nature of property wrappers.

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