Reputation: 104065
I would like to represent a generic JSON object in Swift:
let foo: [String: Any] = [
"foo": 1,
"bar": "baz",
]
But the [String: Any]
type suggested by the compiler doesn’t really work well. I can’t check two instances of the type for equality, for example, while that should be possible with two JSON trees.
What also doesn’t work is using the Codable
machinery to encode that value into a JSON string:
let encoded = try JSONEncoder().encode(foo)
Which blows up with an error:
fatal error: Dictionary<String, Any> does not conform to Encodable because Any does not conform to Encodable.
I know I can introduce a precise type, but I am after a generic JSON structure. I even tried to introduce a specific type for generic JSON:
enum JSON {
case string(String)
case number(Float)
case object([String:JSON])
case array([JSON])
case bool(Bool)
case null
}
But when implementing Codable
for this enum I don’t know how to implement encode(to:)
, since a keyed container (for encoding objects) requires a particular CodingKey
argument and I don’t know how to get that.
Is it really not possible to create an Equatable
generic JSON tree and encode it using Codable
?
Upvotes: 6
Views: 4078
Reputation: 115
I ran into this issue but I had too many types that I want to desserialize so I think I would have to enumerate all of them in the JSON enum from the accepted answer. So I created a simple wraper which worked surprisingly well:
struct Wrapper: Encodable {
let value: Encodable
func encode(to encoder: Encoder) throws {
try value.encode(to: encoder)
}
}
then you could write
let foo: [String: Wrapper] = [
"foo": Wrapper(value: 1),
"bar": Wrapper(value: "baz"),
]
let encoded = try JSONEncoder().encode(foo) // now this works
Not the prettiest code ever, but worked for any type you want to encode without any additional code.
Upvotes: 2
Reputation: 93
You could try this BeyovaJSON
import BeyovaJSON
let foo: JToken = [
"foo": 1,
"bar": "baz",
]
let encoded = try JSONEncoder().encode(foo)
Upvotes: 1
Reputation: 10776
We will use generic strings as coding keys:
extension String: CodingKey {
public init?(stringValue: String) {
self = stringValue
}
public var stringValue: String {
return self
}
public init?(intValue: Int) {
return nil
}
public var intValue: Int? {
return nil
}
}
The rest really is just a matter of getting the correct type of container and writing your values to it.
extension JSON: Encodable {
public func encode(to encoder: Encoder) throws {
switch self {
case .string(let string):
var container = encoder.singleValueContainer()
try container.encode(string)
case .number(let number):
var container = encoder.singleValueContainer()
try container.encode(number)
case .object(let object):
var container = encoder.container(keyedBy: String.self)
for (key, value) in object {
try container.encode(value, forKey: key)
}
case .array(let array):
var container = encoder.unkeyedContainer()
for value in array {
try container.encode(value)
}
case .bool(let bool):
var container = encoder.singleValueContainer()
try container.encode(bool)
case .null:
var container = encoder.singleValueContainer()
try container.encodeNil()
}
}
}
Given this, I'm sure you can implement Decodable
and Equatable
yourself.
Note that this will crash if you try to encode anything other than an array or an object as a top-level element.
Upvotes: 6
Reputation: 15512
From my point of view the most suitable is SwiftyJSON. It has nice API for the developers to be sure how parse and work with JSON
objects.
JSON
object have quite nice interface for working with different types of the response.
From the Apple docs:
Types that conform to the Equatable protocol can be compared for equality using the equal-to operator (==) or inequality using the not-equal-to operator (!=). Most basic types in the Swift standard library conform to Equatable.
Let consider case that you ask.We should just check if JSON
conforms to protocol Equatable
.
Upvotes: -1
Reputation: 1131
You could use generics for this:
typealias JSON<T: Any> = [String: T] where T: Equatable
Upvotes: 2