0
[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;
}
這是json數組imson receving,如何爲此編寫json模式類?前三個沒有key和它的json數組。如何爲此傳入的json格式創建json pojo
前3個參數需要一個名稱。除此之外,它應該是直截了當的。 –
前三個沒有鍵和它的json數組格式。我已經爲其餘參數創建pojo,但不知道如何調用這個對象 – mahe
@ user7118237你是在談論json模式還是java pojo類?我標記了jsonschema,因爲您在問題中提到了這個問題,但現在我不太確定 – eis