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解码对象时枚举值不存在的情况应该如何处理？例如，我将如何处理下面示例中评级类型为“可怕”的情况？如果不存在任何值，有没有办法设置默认值？

public struct Review: Decodable {

    public var ratingType: RatingType?

    enum CodingKeys: String, CodingKey {
        case ratingType = "rating_type"
    }

    public init(from decoder: Decoder) throws {

        let container = try decoder.container(keyedBy: CodingKeys.self)
        ratingType = try container.decodeIfPresent(RatingType.self, forKey: .ratingType)
    }
}

public enum RatingType: String, Codable {
    case Good = "good"
    case Bad = "bad"
}
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2 回答 2

2

decodeIfPresent如果 key 不存在则返回nil，如果 value 无效，它可能会抛出错误。通过使用try?，我们可以这样做：

ratingType = 
    (try? (container.decodeIfPresent(RatingType.self, forKey: .ratingType) ?? <some default value>)
    ) ?? <some default value>
于 2018-01-09T17:07:01.040 回答
1

如果应该实际使用收到的意外评级（针对客户端和服务器之间的版本不匹配的防御性编码或其他），您可以尝试使用不同的枚举定义：

enum RatingType
{
    case good
    case bad
    case other(String)
}

您失去了自动编码，但获得了灵活性。

一个可能的实现：

import Foundation

struct Review
{
    public var rating: RatingType?

    enum CodingKeys: String, CodingKey {
        case rating = "rating_type"
    }

    public init(rating: RatingType) {
        self.rating = rating
    }
}

extension Review: Decodable
{
    public init(from decoder: Decoder) throws {

        let container = try decoder.container(keyedBy: CodingKeys.self)
        if let label = try container.decodeIfPresent(String.self, forKey: .rating) {
            rating = RatingType(label: label)
        }
    }
}

extension Review: Encodable
{
    func encode(to encoder: Encoder) throws {
        var container = encoder.container(keyedBy: CodingKeys.self)
        if let rating = rating {
            try container.encode(rating.label, forKey: .rating)
        }
    }
}

.

enum RatingType
{
    case good
    case bad
    case other(String)

    init(label: String) {
        switch label {
            case "good": self = .good
            case "bad": self = .bad
            default: self = .other(label)
        }
    }

    var label: String {
        switch self {
            case .bad: return "bad"
            case .good: return "good"
            case let .other(label): return label
        }
    }
}

在现实生活中，您可能希望RatingType遵循Codable/Decodable来简化编码/编码。随意修改。

执行编码/解码的示例：

func encode_review(_ review: Review) throws -> String
{
    let data = try JSONEncoder().encode(review)
    return String(data: data, encoding: String.Encoding.utf8) ?? "/* ERROR */"
}

func decode_json(_ json: String) throws -> Review?
{
    guard let data = json.data(using: String.Encoding.utf8) else { return nil }
    let review = try JSONDecoder().decode(Review.self, from: data)
    return review
}

print(try! encode_review(Review(rating: .good)))  // {"rating_type":"good"}
print(try! encode_review(Review(rating: .other("horrible")))) // {"rating_type":"horrible"}

let good_review_json = """
    {"rating_type":"good"}
"""
let great_review_json = """
    {"rating_type":"great"}
"""

if let review = try! decode_json(good_review_json), let label = review.rating?.label {
    print(label) // good
}
if let review = try! decode_json(great_review_json), let label = review.rating?.label {
    print(label) // great
}
于 2018-01-09T17:56:22.243 回答