java在Spring Boot应用程序的JSON中包含嵌套对象

Question

有谁能指导我如何将JSON对象激活嵌套在JSON对象数据下

请参见下面的myJSON：

{
    "status": "Success",
    "DATA": {
        "aid": "0",
        "app": "OASIS_EDGE",
        "actkey": "STAR2020",
        "expiredate": "2025-08-05 04:40:28.0",
        "partner": "X",
        "nsp": "NAD"
    }
}

JSON的JSON我实际上期望：

{
    "STATUS": "SUCCESS",
    "DATA": {
        "ACTIVATION": {
            "aid": "0",
            "app": "OASIS_EDGE",
            "actkey": "STAR2020",
            "expiredate": "2025-08-05 04:40:28.0",
            "partner": "X",
            "nsp": "NAD"
        }
    }
}

这是我的激活响应。java类负责创建JSON：

public class ActivationResponse {   
    
    private String Status;
        
    public String getStatus()
    {
        return Status;
    }
    
    public void setStatus(String Status) {
        this.Status = Status;
    }
    
    @JsonProperty("DATA")   
        
    private ActivationModel activationData;
     
    
    public ActivationModel getActivationData() {
        return activationData;
    }

    public void setActivationData(ActivationModel activationData2) {
        this.activationData = activationData2;
    }

}

Answer 1

1. 使用com中的JsonNode。fasterxml。杰克逊。数据绑定。JsonNodea la:

import com.fasterxml.jackson.databind.JsonNode;

public class ActivationResponse {
    
    private String Status;

    private JsonNode activationData;
        
    public String getStatus() return Status;
    
    public void setStatus(String Status) this.Status = Status;  
    
    public ActivationModel getActivationData() return activationData;

    public void setActivationData(ActivationModel activationData2) {
        activationData = activationData2;
    }
}

1. 使用来自java的好的“ol映射”。util。映射像这样：

import java.util.Map;

public class ActivationResponse {
    
    private String Status;

    private Map<String, Object> activationData;
        
    public String getStatus() return Status;
    
    public void setStatus(String Status) this.Status = Status;  
    
    public ActivationModel getActivationData() return activationData;

    public void setActivationData(ActivationModel activationData2) {
        activationData = activationData2;
    }
}

我无耻地借用了这个tutorial（没有附属关系）

Answer 2

由于需要在数据下激活，因此需要一个带有激活字段的类数据，如下面的示例所示

class MyJson {
  private String Status;

  @JsonProperty("DATA")   
  private DataModel data;
};
 
class DataModel {
  @JsonProperty("ACTIVATION")   
  private ActivationModel activation;
};