问题
Swift 4 added the new Codable protocol. When I use JSONDecoder it seems to require all the non-optional properties of my Codable class to have keys in the JSON or it throws an error.
Making every property of my class optional seems like an unnecessary hassle since what I really want is to use the value in the json or a default value. (I don't want the property to be nil.)
Is there a way to do this?
class MyCodable: Codable {
var name: String = "Default Appleseed"
}
func load(input: String) {
do {
if let data = input.data(using: .utf8) {
let result = try JSONDecoder().decode(MyCodable.self, from: data)
print("name: \(result.name)")
}
} catch {
print("error: \(error)")
// `Error message: "Key not found when expecting non-optional type
// String for coding key \"name\""`
}
}
let goodInput = "{\"name\": \"Jonny Appleseed\" }"
let badInput = "{}"
load(input: goodInput) // works, `name` is Jonny Applessed
load(input: badInput) // breaks, `name` required since property is non-optional
回答1:
Approach that I prefer is using so called DTOs - data transfer object. It is a struct, that conforms to Codable and represents the desired object.
struct MyClassDTO: Codable {
let items: [String]?
let otherVar: Int?
}
Then you simply init the object that you want to use in the app with that DTO.
class MyClass {
let items: [String]
var otherVar = 3
init(_ dto: MyClassDTO) {
items = dto.items ?? [String]()
otherVar = dto.otherVar ?? 3
}
var dto: MyClassDTO {
return MyClassDTO(items: items, otherVar: otherVar)
}
}
This approach is also good since you can rename and change final object however you wish to. It is clear and requires less code than manual decoding. Moreover, with this approach you can separate networking layer from other app.
回答2:
You can implement the init(from decoder: Decoder) method in your type instead of using the default implementation:
class MyCodable: Codable {
var name: String = "Default Appleseed"
required init(from decoder: Decoder) throws {
let container = try decoder.container(keyedBy: CodingKeys.self)
if let name = try container.decodeIfPresent(String.self, forKey: .name) {
self.name = name
}
}
}
You can also make name a constant property (if you want to):
class MyCodable: Codable {
let name: String
required init(from decoder: Decoder) throws {
let container = try decoder.container(keyedBy: CodingKeys.self)
if let name = try container.decodeIfPresent(String.self, forKey: .name) {
self.name = name
} else {
self.name = "Default Appleseed"
}
}
}
or
required init(from decoder: Decoder) throws {
let container = try decoder.container(keyedBy: CodingKeys.self)
self.name = try container.decodeIfPresent(String.self, forKey: .name) ?? "Default Appleseed"
}
Re your comment: With a custom extension
extension KeyedDecodingContainer {
func decodeWrapper<T>(key: K, defaultValue: T) throws -> T
where T : Decodable {
return try decodeIfPresent(T.self, forKey: key) ?? defaultValue
}
}
you could implement the init method as
required init(from decoder: Decoder) throws {
let container = try decoder.container(keyedBy: CodingKeys.self)
self.name = try container.decodeWrapper(key: .name, defaultValue: "Default Appleseed")
}
but that is not much shorter than
self.name = try container.decodeIfPresent(String.self, forKey: .name) ?? "Default Appleseed"
回答3:
One solution would be to use a computed property that defaults to the desired value if the JSON key is not found. This adds some extra verbosity as you'll need to declare another property, and will require adding the CodingKeys enum (if not already there). The advantage is that you don't need to write custom decoding/encoding code.
For example:
class MyCodable: Codable {
var name: String { return _name ?? "Default Appleseed" }
var age: Int?
private var _name: String?
enum CodingKeys: String, CodingKey {
case _name = "name"
case age
}
}
回答4:
You can implement.
struct Source : Codable {
let id : String?
let name : String?
enum CodingKeys: String, CodingKey {
case id = "id"
case name = "name"
}
init(from decoder: Decoder) throws {
let values = try decoder.container(keyedBy: CodingKeys.self)
id = try values.decodeIfPresent(String.self, forKey: .id) ?? ""
name = try values.decodeIfPresent(String.self, forKey: .name)
}
}
回答5:
If you don't want to implement your encoding and decoding methods, there is somewhat dirty solution around default values.
You can declare your new field as implicitly unwrapped optional and check if it's nil after decoding and set a default value.
I tested this only with PropertyListEncoder, but I think JSONDecoder works the same way.
回答6:
If you think that writing your own version of init(from decoder: Decoder) is overwhelming, I would advice you to implement a method which will check the input before sending it to decoder. That way you'll have a place where you can check for fields absence and set your own default values.
For example:
final class CodableModel: Codable
{
static func customDecode(_ obj: [String: Any]) -> CodableModel?
{
var validatedDict = obj
let someField = validatedDict[CodingKeys.someField.stringValue] ?? false
validatedDict[CodingKeys.someField.stringValue] = someField
guard
let data = try? JSONSerialization.data(withJSONObject: validatedDict, options: .prettyPrinted),
let model = try? CodableModel.decoder.decode(CodableModel.self, from: data) else {
return nil
}
return model
}
//your coding keys, properties, etc.
}
And in order to init an object from json, instead of:
do {
let data = try JSONSerialization.data(withJSONObject: json, options: .prettyPrinted)
let model = try CodableModel.decoder.decode(CodableModel.self, from: data)
} catch {
assertionFailure(error.localizedDescription)
}
Init will look like this:
if let vuvVideoFile = PublicVideoFile.customDecode($0) {
videos.append(vuvVideoFile)
}
In this particular situation I prefer to deal with optionals but if you have a different opinion, you can make your customDecode(:) method throwable
来源：https://stackoverflow.com/questions/44575293/with-jsondecoder-in-swift-4-can-missing-keys-use-a-default-value-instead-of-hav