如何在 `codable` 响应中使用字典？

Question

我从 JSON 查询中得到以下响应。我如何将字典表示为可编码的？我已缩短 JSON 响应以保存 space。

{
 "result":[
            {
            "delivery_address": "",
            "made_sla": "true",
            "watch_list": "",
            "upon_reject": "cancel",
            "location": {
                "link": "https://foo.com/api/now/table/cmn_location/753ac76f4fd9320066bfb63ca310c79b",
                "value": "753ac76f4fd9320066bfb63ca310c79b"
             }
          }
    ]
}    

struct ResultList : Codable {
   let result: [Result]
}

struct Result : Codable {
    let delivery_address: String
    let made_sla: String
    let watch_list: String
    let upon_reject: String   
    let location: Location
}

struct Location: Codable {
    let link: String?
    let value: String?
}

  let decoder = JSONDecoder()
  do {
      let todo = try decoder.decode(ResultList.self, from: responseData)

      print("todo \(todo)")

      } catch {
        print("error trying to convert data to JSON")
        print(error)

      }

我收到以下错误：

"Expected to decode Dictionary<String, Any> but found a string/data instead.", underlyingError: nil))

Answer 1

假设您的 JSON 是（请注意缺少的右括号）

{
"result": [
    {
     "delivery_address": "",
     "made_sla": "true",
     "watch_list": "",
     "upon_reject": "cancel",
     "location": {
                "link": "https://blah/blah/foo",
                "value": "fsfdfr32r34rwfwffas"
     }
    }
  ]
}

你可以解码这些结构

struct Root : Decodable {
    let result : [Result]

    struct Result : Decodable {
        let location: Location
    }
}

struct Location: Decodable {
    let link: String
    let value: String
}

和

JSONDecoder().decode(Root.self, from: data)

Answer 2

根据所有评论，我相信 JSON 你实际解码的看起来更像这样：

{
    "result": [{
            "delivery_address": "",
            "made_sla": "true",
            "watch_list": "",
            "upon_reject": "cancel",
            "location": {
                "link": "https://foo.com/api/now/table/cmn_location/753ac76f4fd9320066bfb63ca310c79b",
                "value": "753ac76f4fd9320066bfb63ca310c79b"
            }
        },
        {
            "delivery_address": "",
            "made_sla": "true",
            "watch_list": "",
            "upon_reject": "cancel",
            "location": ""

        }
    ]
}

所以有些记录有位置，有些记录将位置编码为空字符串。基本上这在很多层面上确实是一团糟JSON，无论生成什么代码都应该修复。坏消息我相信那不会发生。好消息是我们至少可以在本地修复它。

我们将不得不手动解码它，所以我们不妨在我们在这里的时候清理所有其他的烂摊子。第一件事是名称不符合 Swift 命名约定。我们可以使用 CodingKeys 来解决这个问题。我们还可以为当前输入错误的字符串提供真实类型（Bool 和 Rejection 枚举）。

enum Rejection: String, Codable {
    case cancel
}
struct Result : Codable {
    let deliveryAddress: String
    let madeSLA: Bool
    let watchList: String
    let uponReject: Rejection
    let location: Location?

    private enum CodingKeys: String, CodingKey {
        case deliveryAddress = "delivery_address"
        case madeSLA = "made_sla"
        case watchList = "watch_list"
        case uponReject = "upon_reject"
        case location
    }
}

现在我们只需要能够解码它。请注意，我将 Location 设为可选。显然它有时不存在，所以您要么需要一个默认值，要么它需要是可选的。我选择了后者。解码所有这些非常简单：

init(from decoder: Decoder) throws {
    let container = try decoder.container(keyedBy: CodingKeys.self)
    deliveryAddress = try container.decode(String.self, forKey: .deliveryAddress)
    madeSLA = try container.decode(String.self, forKey: .madeSLA) == "true"
    watchList = try container.decode(String.self, forKey: .watchList)
    uponReject = try container.decode(Rejection.self, forKey: .uponReject)

    location = try? container.decode(Location.self, forKey: .location)
}

最后一行是您的实际问题。它只是说如果我们不能将它解码为 Location，则将其设置为 nil。我们可以在这里更严格，首先尝试将其解码为 Location，然后解码为 String，然后检查 String 是否为空，但是对于任何解码失败，在这里使用 nil 感觉是合理的。

如何在 `codable` 响应中使用字典？

How to use a dictionary in `codable` response?

ios

json

swift

codable