Question

我从JSON查询得到以下响应。如何将字典表示为可编码字典？我缩短了JSON响应以节省空间。

{
 "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

基于所有评论，我相信您实际解码的JSON看起来更像是这样：

{
    "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": ""

        }
    ]
}

因此，有些记录有一个位置，有些记录将位置编码为空字符串。基本上这是在很多级别上搞砸了JSON，无论代码生成什么都应该修复。坏消息我肯定不会发生。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，所以当我们在这里时，我们不妨清理所有其他的混乱。第一件事是名称与Swift命名约定不匹配。我们可以使用CodingKeys来解决这个问题。我们还可以为当前错误输入的字符串提供真实类型（Bool和拒绝枚举）。

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设置为可选。显然，它有时不存在，因此您需要一个默认值，或者它需要是可选的。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)
}

最后一行是你的实际问题。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并首先尝试将其解码为Location，然后作为String解析，然后检查String是否为空，但是在这里使用nil解决任何解码失败都是合理的。

Answer 2

假设您的JSON是（请注意缺少的右括号）

struct Root : Decodable {
    let result : [Result]

    struct Result : Decodable {
        let location: Location
    }
}

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

您可以解码这些结构

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

带

{{1}}

如何在`codable`响应中使用字典？

2 个答案: