使用 [JsonPropertyName] 标签解析嵌套到 JSON 文本中的值

Question

使用 C#、Newtonsoft.Json 13.0.2 并具有以下可能的 JSON：

Json 类型 1 ---------------------

{
  "type": "succeeded",
  "event_date": "2023",
  "transaction": {
    "id": "AAA",
    "authorization": "123",
    "operation_type": "456"
  }
}

Json 类型 2 ---------------------

{
  "type": "failed",
  "event_date": "2023",
  "failureReport": {
    "id": "AAA",
    "failureType": "123"      
  }
}

我使用以下代码来反序列化两个对象中的任何一个，如下所示：

[Table("MySql_Test_Table")]

public class TestClass
{
[Key]
[JsonPropertyName("id")]
public int id { get; set; }

[JsonPropertyName("event_date")]
public DateTime? event_date { get; set; }

[JsonPropertyName("type")]
public String? type { get; set; }

[JsonPropertyName("transaction.id")]
public String? transaction_id { get; set; }  

[JsonPropertyName("transaction.authorization")]
public String? authorization{ get; set; }

[JsonPropertyName("transaction.operation_type")]
public String? operation_type{ get; set; }

[JsonPropertyName("failureReport.id")]
public String? failureReportId{ get; set; }

[JsonPropertyName("failureReport.failureType")]
public String? failureReportType{ get; set; }

}

TestClass testClass = Newtonsoft.Json.JsonConvert.DeserializeObject<TestClass>(oneOfTwoPossiblejsonTexts.ToString());

//...Code for saving testClass using EntityFramrwork...

使用上述代码，我可以毫无问题地获取 type 和 event_date 的值。但是我无法填充 id、授权、操作类型、failureReportId 或failureReportType。 我想问题是我的 [JsonPropertyName] 标签。请问，我该如何更正它以访问所需的值？

Answer 1

如果你想将这些 json 反序列化到你的 Flatten 类中，你可以创建一个 JsonConstructor

public class TestClass
{
    public int id { get; set; }

    [JsonProperty("event_date")]
    public string? eventDate { get; set; }

    public String? type { get; set; }

    public String? transactionId { get; set; }

    public String? authorization { get; set; }

    public String? operationType { get; set; }

    public String? failureReportId { get; set; }

    public String? failureReportType { get; set; }

    [JsonConstructor]
    public TestClass(JToken transaction, JToken failureReport)
    {
        if (transaction != null)
        {
            transactionId = (string?)transaction["id"];
            authorization = (string?)transaction["authorization"];
            operationType = (string?)transaction["operation_type"];
        };
        if (failureReport != null)
        {
            failureReportId = (string?)failureReport["id"];
            failureReportType = (string?)failureReport["failureType"];
        };
    }

     public TestClass() { }
}

由于您使用 Newtonsoft.Json 进行反序列化，因此必须使用 [JsonProperty] 而不是 [JsonPropertyName]

Answer 2

最安全的选择可能是创建相应的 C# 类型：

public class TestClass
{
[Key]
[JsonPropertyName("id")]
public int id { get; set; }

[JsonPropertyName("event_date")]
public DateTime? event_date { get; set; }

[JsonPropertyName("type")]
public String? type { get; set; }

public TransactionType Transaction { get; set; }

public FailureReportType FailureReport { get; set; }
}

public class TransactionType
{
[JsonPropertyName("id")]
public String? transaction_id { get; set; }  

[JsonPropertyName("authorization")]
public String? authorization{ get; set; }

[JsonPropertyName("operation_type")]
public String? operation_type{ get; set; }
}

public class FailureReportType
{
[JsonPropertyName("id")]
public String? failureReportId{ get; set; }

[JsonPropertyName("failureType")]
public String? failureReportType{ get; set; }
}

Answer 3

在我的例子中，我必须获得

name.givenName

和

name.familyName

嵌套属性，我发现这个解决方案对于

System.Text.Json

来说是最简单的（无需为

Name

创建额外的类）：

[JsonPropertyName("name")]
public Dictionary<string, string> Name { get; set; }

public string FirstName { get => Name["givenName"]; }
public string LastName { get => Name["familyName"]; }