如何为这个传入的json格式创建json pojo

时间:2017-06-22 10:15:12

标签: json

[2,"1498134496","StatusNotification",{"connectorId":"1","errorCode":"NoError","info":"NoError","status":"Available","timestamp": "2017-06-22 12:28:16","vendorId":"CPV07","vendorErrorCode":"123ASD"}]
package com.chakra.ev.webservice.jsonschema;
import javax.persistence.Column;
import lombok.AllArgsConstructor;
import lombok.Getter;
import lombok.RequiredArgsConstructor;
import lombok.Setter;
import lombok.ToString;
import org.codehaus.jackson.annotate.JsonIgnoreProperties;
@Getter
@Setter 
@AllArgsConstructor
@ToString
@RequiredArgsConstructor
@JsonIgnoreProperties(ignoreUnknown = true)
public class StatusNotificationSchema {
   public Integer connectorId;
    public String errorCode;
    public String info;
    public String status;
    public String timestamp;
    public String vendorId;
    public String vendorErrorCode;
}

这是我接收的json数组,如何为此编写json schema类?前三个没有密钥及其json数组。

1 个答案:

答案 0 :(得分:0)

我认为主要问题是前三个数组参数。这是有效的JSON。我认为json架构应该是这样的:

{
  "type": "array",
  "items": [
    {
      "type": "number"
    },
    {
      "type": "string"
    },
    {
      "type": "string",
      "enum": ["StatusNotification", ...]
    },
    {
      "type": "object",
      "properties": {
         "connectionId": {
             //...
         },  //...
      }, // ...
    }
  ]
}