用于Spring Boot的Rest API和Web服务

Question

我有春季启动应用程序，使用jdbctemplate可以使用此URL显示数据

http://localhost:8080/query

结果像这样

[{"id_data":1,"id_user":1,"time":"Thursday, April 09, 2015 18:09:26","ecgvalue":3.3871,"inputtime":"2015-04-09 18:11:25.0"},{"id_data":2,"id_user":1,"time":"Thursday, April 09, 2015 18:09:26","ecgvalue":1.56892,"inputtime":"2015-04-09 18:11:25.0"},{"id_data":3,"id_user":1,"time":"Thursday, April 09, 2015 18:09:26","ecgvalue":1.60802,"inputtime":"2015-04-09 18:11:26.0"},{"id_data":4,"id_user":1,"time":"Thursday, April 09, 2015 18:09:26","ecgvalue":2.09677,"inputtime":"2015-04-09 18:11:26.0"},{"id_data":5,"id_user":1,"time":"Thursday, April 09, 2015 18:09:26","ecgvalue":1.99902,"inputtime":"2015-04-09 18:11:26.0"}]

我知道这是json对象。 我的问题是如何从中制作Web服务？ 像Rest Web服务一样，这是我的代码QueryController.java

package com.ewsn.eepiscure.controller;

/**
 *
 * @author sammy
 */
import java.util.List;
import javax.sql.DataSource;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.jdbc.core.JdbcTemplate;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RestController;

@RestController
public class QueryController {

    @Autowired
    protected JdbcTemplate hiveTemplate;

    @RequestMapping("/query")

    public List query() {
        List data = hiveTemplate.queryForList("select * from ecg.hivetbluserdata limit 100");
        return data;
    }


}

Answer 1

如果您对使用Spring在Java中创建“ REST Web服务”感兴趣，建议您看一下Spring Data Rest模块，该模块可用于创建

在Spring Data存储库之上的超媒体驱动的REST Web服务

见http://projects.spring.io/spring-data-rest/