Question

我从YAML文件生成快速模型时遇到问题。

swagger-codegen generate -i swagger_1.yaml -l swift4

下面的

是示例模型之一。由于没有编码键

，因此无法快速构建。

import Foundation

open class User: Codable {

    public enum Sex: String, Codable { 
        case male = "male"
        case female = "female"
    }
    public var id: Int64?
    public var username: String?
    public var firstName: String?
    public var lastName: String?
    public var sex: Sex?



    public init(id: Int64?, username: String?, firstName: String?, lastName: String?, sex: Sex?) {
        self.id = id
        self.username = username
        self.firstName = firstName
        self.lastName = lastName
        self.sex = sex
    }


    // Encodable protocol methods

    public func encode(to encoder: Encoder) throws {

        var container = encoder.container(keyedBy: String.self)

        try container.encodeIfPresent(id, forKey: "id")
        try container.encodeIfPresent(username, forKey: "username")
        try container.encodeIfPresent(firstName, forKey: "firstName")
        try container.encodeIfPresent(lastName, forKey: "lastName")
        try container.encodeIfPresent(sex, forKey: "sex")
    }

    // Decodable protocol methods

    public required init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: String.self)

        id = try container.decodeIfPresent(Int64.self, forKey: "id")
        username = try container.decodeIfPresent(String.self, forKey: "username")
        firstName = try container.decodeIfPresent(String.self, forKey: "firstName")
        lastName = try container.decodeIfPresent(String.self, forKey: "lastName")
        sex = try container.decodeIfPresent(Sex.self, forKey: "sex")
    }
}

也在GitHub上发布了这个问题。

https://github.com/swagger-api/swagger-codegen/issues/8289

Answer 1

您需要添加enum CodingKeys并重写initFromDecoder / encode。

同样，在您的情况下，根本不需要单独的编码/解码/编码键，因为类的所有属性均已可编码。

open class User: Codable {
    ....
    enum CodingKeys: String, CodingKey {
        case id, username, firstName, lastName, sex
    }

    // Encodable protocol methods

    public func encode(to encoder: Encoder) throws {

        var container = encoder.container(keyedBy: CodingKeys.self)

        try container.encodeIfPresent(id, forKey: .id)
        try container.encodeIfPresent(username, forKey: .username)
        try container.encodeIfPresent(firstName, forKey: .firstName)
        try container.encodeIfPresent(lastName, forKey: .lastName)
        try container.encodeIfPresent(sex, forKey: .sex)
    }

    // Decodable protocol methods

    public required init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)

        id = try container.decodeIfPresent(Int64.self, forKey: .id)
        username = try container.decodeIfPresent(String.self, forKey: .username)
        firstName = try container.decodeIfPresent(String.self, forKey: .firstName)
        lastName = try container.decodeIfPresent(String.self, forKey: .lastName)
        sex = try container.decodeIfPresent(Sex.self, forKey: .sex)
    }
}

Swift 4支持swagger代码生成

1 个答案: