可解码符合枚举类型的属性

Question

我有这个枚举:

enum DealStatus:String {
    case PENDING = "Pending"
    case ACTIVE = "Active"
    case STOP = "Stop"
    case DECLINED = "Declined"
    case PAUSED = "Paused"
}

和结构:

struct ActiveDeals: Decodable {
    let keyword:            String
    let bookingType:        String
    let expiryDate:         Int
    let createdAt:          Int?
    let shopLocation:       String?
    let dealImages:         [DealImages]?
    let dealStatus:         String?
    let startingDate:       Int?
}

在结构中我试图将枚举作为类型分配dealStatus如下:

struct ActiveDeals: Decodable {
        let keyword:            String
        let bookingType:        String
        let expiryDate:         Int
        let createdAt:          Int?
        let shopLocation:       String?
        let dealImages:         [DealImages]?
        let dealStatus:         DealStatus
        let startingDate:       Int?
    }

但我收到一些编译器错误:

类型'ActiveDeals'不符合协议'可解码'

协议需要初始化程序'init(from :)',类​​型为'Decodable'(Swift.Decodable)

无法自动合成'Decodable',因为'DealStatus'不符合'Decodable'

Answer 1

问题是Swift可以自动合成所需的方法,Decodable只有当结构的所有属性都是,Decodable而你的枚举不是Decodable.

刚刚在游乐场玩了一下,看起来你可以Decodable 通过声明它来制作你的枚举,并且Swift会自动为你合成这些方法.即

enum DealStatus:String, Decodable  
//                      ^^^^^^^^^ This is all you need
{
    case PENDING = "Pending"
    case ACTIVE = "Active"
    case STOP = "Stop"
    case DECLINED = "Declined"
    case PAUSED = "Paused"
}