第一件事：我想从api读取json数据，并想在我的thymeleaf模板页面中显示。以下是我的代码：
我的json类：
package com.mslapiagent.entity;
import java.math.BigInteger;
import javax.persistence.Entity;
public class MSLApiAgent{
private int id;
private BigInteger tranId;
private String clientTranId;
private String msisdn;
private String msgbody;
public MSLApiAgent() {
}
public MSLApiAgent(int id, BigInteger tranId, String clientTranId, String msisdn, String msgbody) {
this.id = id;
this.tranId = tranId;
this.clientTranId = clientTranId;
this.msisdn = msisdn;
this.msgbody = msgbody;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public BigInteger getTranId() {
return tranId;
}
public void setTranId(BigInteger tranId) {
this.tranId = tranId;
}
public String getClientTranId() {
return clientTranId;
}
public void setClientTranId(String clientTranId) {
this.clientTranId = clientTranId;
}
public String getMsisdn() {
return msisdn;
}
public void setMsisdn(String msisdn) {
this.msisdn = msisdn;
}
public String getMsgbody() {
return msgbody;
}
public void setMsgbody(String msgbody) {
this.msgbody = msgbody;
}
@Override
public String toString() {
return "MSLApiAgent [id=" + id + ", tranId=" + tranId + ", clientTranId=" + clientTranId + ", msisdn=" + msisdn
+ ", msgbody=" + msgbody + "]";
}
}我的控制器：
@Controller
public class ApiAgentController {
@RequestMapping("/test02")
public String attaComsianTest02(Model model, MSLApiAgent mslApiAgent) {
// request url
String url = "http://localhost:8080/MSLSystem_3/api/v1/messages/4";
// create an instance of RestTemplate
RestTemplate restTemplate2 = new RestTemplate();
// make an HTTP GET request
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_JSON);
headers.setAccept(Collections.singletonList(MediaType.APPLICATION_JSON));
HttpEntity request = new HttpEntity(headers);
ResponseEntity<MSLApiAgent> exchange = restTemplate2.exchange(url, HttpMethod.GET,request,MSLApiAgent.class);
MSLApiAgent body = exchange.getBody();
int id = body.getId();
BigInteger tranId = body.getTranId();
String clientTranId = body.getClientTranId();
String msisdn = body.getMsisdn();
String msgbody = body.getMsgbody();
model.addAttribute("exchanges", body);
return "test02";
}
}最后是我的查看页面：
<!DOCTYPE html>
<html lang="en" xmlns:th="http://www.w3.org/1999/xhtml">
<head>
<meta charset="UTF-8">
<title>test01</title>
</head>
<body>
<h1>Here is your API data</h1>
<table>
<thead>
<tr>
<td>Id</td>
<td>Transaction Id</td>
<td>Client Id</td>
<td>Mobile No.</td>
<td>Messages</td>
</tr>
</thead>
<tbody>
<tr data-th-each="exchange : ${exchanges}">
<td data-th-text="${exchange.id}">...</td>
<td data-th-text="${exchange.tranId}">...</td>
<td data-th-text="${exchange.clientTranId}">...</td>
<td data-th-text="${exchange.msisdn}">...</td>
<td data-th-text="${exchange.msgbody}">...</td>
</tr>
</tbody>
</table>
</body>
</html>当我选择一个像
String url = "http://localhost:8080/MSLSystem_3/api/v1/messages/4";然后它就出现在我的视野里。但当我使用
String url = "http://localhost:8080/MSLSystem_3/api/v1/messages";现在它向我显示任何数据，我得到一个错误：
路径为[]的上下文中servlet[dispatcherservlet]的servlet.service（）引发异常[请求处理失败；嵌套的异常是org.springframework.web.client.restclientexception:提取类型[class com.mslapiagent.entity.mslapiagent]和内容类型[application/json]的响应时出错；嵌套异常为org.springframework.http.converter.httpMessageNotradableException:json分析错误：无法反序列化的示例 com.mslapiagent.entity.MSLApiAgent 启动外\u数组令牌；嵌套异常为com.fasterxml.jackson.databind.exc.missmatchdinputException:无法反序列化的示例 com.mslapiagent.entity.MSLApiAgent [source:（pushbackinputstream）处的启动\u数组令牌不足；行：1，列：1]]具有根本原因com.fasterxml.jackson.databind.exc.missmatchdinputException:无法反序列化的示例 com.mslapiagent.entity.MSLApiAgent [source:（pushbackinputstream）处的启动\u数组令牌不足；行：1，列：1]
我该怎么办？
第二件事：事实上，我想检查这个数据的随机基础，就像1秒间隔计划和存储最后的记录到我的数据库。好的方法是什么？
我已经尝试使用timer和timertask准备api url。但我失败了。
1条答案
按热度按时间fdbelqdn1#
当你使用
http://localhost:8080/MSLSystem_3/api/v1/messages，它很可能返回一个json对象数组。无法将其序列化到
MSLApiAgent在这里上课：您应该创建一个表示
MSLApiAgent物体：然后使用：