Question

我一直在尝试使用alamofire从url获得json响应。创建了模型，apirouter和api客户端类。

显示错误

 failure(Alamofire.AFError.responseSerializationFailed(reason: 
  Alamofire.AFError.ResponseSerializationFailureReason.decodingFailed(error: 
   Swift.DecodingError.keyNotFound(CodingKeys(stringValue: "variables", intValue: nil), Swift.DecodingError.Context(codingPath: [_JSONKey(stringValue: "Index 0", intValue: 0)], debugDescription: "No value associated with key CodingKeys(stringValue: \"variables\", intValue: nil) (\"variables\").", underlyingError: nil)))))

这是我的邮递员json回复：

   [
{
    "id": "00602c70-fc8a-11e9-ad1d-2abe2670111d",
    "resourceType": "Task",
    "name": "My Tasks",
    "owner": null,
    "query": {
        "assigneeExpression": "${currentUser()}",
        "taskVariables": [],
        "processVariables": [],
        "caseInstanceVariables": [],
        "orQueries": []
    },
    "properties": {
        "variables": [
            {
                "name": "loanAmount",
                "label": "Loan Amount"
            },
            {
                "name": "firstName",
                "label": "First Name"
            }
        ],
        "color": "#555555",
        "showUndefinedVariable": false,
        "description": "Tasks assigned to me",
        "refresh": false,
        "priority": -10
      }
    }
   ]

我试图从json响应中获取 id，名称和属性->变量->名称和标签的值。

这是模型类别：

  import Foundation

public struct Filter: Codable {

let id: String
let name: String
let properties: [variables]


}

public struct variables: Codable {

   let name: String
   let label: String

}

以下是alamofire的代码：

    private static func performRequest<T:Decodable>(route:APIRouter, decoder: JSONDecoder = JSONDecoder(), completion:@escaping (AFResult<T>)->Void) -> DataRequest {

    return AF.request(route)
                    .responseDecodable (decoder: decoder){ (response: AFDataResponse<T>) in
                        completion(response.result)
                        print("framework response::",response.result)


    }


}


  public static func getFilter(completion:@escaping (AFResult<[Filter]>)->Void) {
       let jsonDecoder = JSONDecoder()
    performRequest(route: APIRouter.getFilter, decoder: jsonDecoder, completion: completion)
   }

任何帮助都非常感谢...

Answer 1

您的模型类应如下所示。

import Foundation

public struct Filter: Codable {

let id: String
let name: String
let properties: Properties


}

public struct Properties: Codable {
   let variables: [variables]
   let color: String
   let showUndefinedVariable: Bool
   let description: String
   let refresh: Bool
   let priority: Int

}

public struct variables: Codable {

   let name: String
   let label: String

}

Answer 2

您尝试过这个吗？这应该是

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  <div id='arrow'></div>
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    <tr>
      <td >0</td>
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      <td id="end">9</td>
    </tr>
    <tr>
      <td>0</td>
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    </tr>
    <tr>
      <td>0</td>
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    </tr>
    <tr>
      <td id="start">0</td>
      <td>1</td>
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    </tr>
  </table>
</div>

Answer 3

您收到的错误消息非常清楚：No value associated with key CodingKeys(stringValue: \"variables\"

您正在尝试将JSON解码为Filter结构，但是JSON没有variables属性。您可以通过引入一个新的Properties结构来解决此问题，该结构将包裹variables属性，如下所示：

struct Filter: Codable {
    let id: String
    let name: String
    let properties: Properties
}

struct Properties: Codable {
    let variables: Variables
}

struct Variables: Codable {
    let name: String
    let label: String
}

此外，正如您在本片段中所看到的，习惯上是在CamelCase中编写类型名称，所以用struct Variables代替struct variables。

如何在iOS Swift中使用Alamofire获得JSON响应？

3 个答案: