在Mysql中的GROUP BY分组，如何取最新的一条记录？

日常开发当中，经常会遇到查询分组数据中最新的一条记录，比如统计当前系统每个人的最新登录记录、外卖系统统计所有买家最新的一次订单记录、图书管理系统借阅者最新借阅书籍的记录等等。今天给大家介绍一下如何实现以上场景的SQL写法，希望对大家能有所帮助!

-- 借阅者表 
CREATE TABLE `userinfo` ( 
  `uid` int(11) NOT NULL AUTO_INCREMENT COMMENT '主键', 
  `uname` varchar(20) NOT NULL COMMENT '姓名', 
  `uage` int(11) NOT NULL COMMENT '年龄', 
  PRIMARY KEY (`uid`) USING BTREE 
) ENGINE=InnoDB AUTO_INCREMENT=8 DEFAULT CHARSET=utf8 ROW_FORMAT=COMPACT; 
INSERT INTO `userinfo` VALUES (1, '小明', 20); 
INSERT INTO `userinfo` VALUES (2, '小张', 30); 
INSERT INTO `userinfo` VALUES (3, '小李', 28); 
-- 书籍表 
CREATE TABLE `bookinfo` ( 
  `id` int(11) NOT NULL AUTO_INCREMENT COMMENT '主键', 
  `book_no` varchar(20) NOT NULL COMMENT '书籍编号', 
  `book_name` varchar(20) NOT NULL COMMENT '书籍名称', 
  PRIMARY KEY (`id`) USING BTREE 
) ENGINE=InnoDB AUTO_INCREMENT=14 DEFAULT CHARSET=utf8 ROW_FORMAT=COMPACT; 
INSERT INTO `bookinfo` VALUES (1, 'ISBN001', '计算机基础'); 
INSERT INTO `bookinfo` VALUES (2, 'ISBN002', '计算机网络'); 
INSERT INTO `bookinfo` VALUES (3, 'ISBN003', '高等数学'); 
INSERT INTO `bookinfo` VALUES (4, 'ISBN004', '明朝那些事'); 
INSERT INTO `bookinfo` VALUES (5, 'ISBN005', '物理'); 
INSERT INTO `bookinfo` VALUES (13, 'ISBN006', '读者'); 
-- 借阅记录表 
CREATE TABLE `borrow_record` ( 
  `id` int(11) NOT NULL AUTO_INCREMENT COMMENT '主键', 
  `user_id` int(11) NOT NULL COMMENT '用户id', 
  `book_id` int(11) NOT NULL COMMENT '书籍id', 
  `borrowtime` datetime NOT NULL COMMENT '书籍id', 
  PRIMARY KEY (`id`) USING BTREE 
) ENGINE=InnoDB AUTO_INCREMENT=16 DEFAULT CHARSET=utf8 ROW_FORMAT=COMPACT; 
INSERT INTO `borrow_record` VALUES (8, 1, 2, '2021-05-01 10:52:00'); 
INSERT INTO `borrow_record` VALUES (9, 2, 4, '2021-07-12 23:32:00'); 
INSERT INTO `borrow_record` VALUES (10, 2, 1, '2021-03-21 09:00:00'); 
INSERT INTO `borrow_record` VALUES (11, 1, 3, '2021-08-11 17:39:00'); 
INSERT INTO `borrow_record` VALUES (12, 1, 5, '2021-09-02 18:12:00'); 
INSERT INTO `borrow_record` VALUES (13, 3, 1, '2021-07-06 12:32:00'); 
INSERT INTO `borrow_record` VALUES (14, 2, 1, '2021-08-09 10:10:00'); 
INSERT INTO `borrow_record` VALUES (15, 4, 3, '2021-04-15 19:45:00' 

写法1 直接group by 根据userid ，使用聚合函数max取得最近的浏览时间

select a.user_id ,max(c.uname) uname 
,max(a.borrowtime) borrowtime,max(b.book_name) book_name 
from borrow_record a  
INNER JOIN bookinfo b on b.id=a.book_id 
INNER JOIN userinfo c on c.uid=a.user_id 
GROUP BY a.user_id 
-- 说明： 这样会存在获取书籍名称错乱的情况， 
-- 因为使用聚合函数获取的书籍名称，不一定是对应用户 
-- 最新浏览记录对应的书籍名称 

写法2 采用子查询的方式，获取借阅记录表最近的浏览时间作为查询条件

select a.user_id ,c.uname,a.borrowtime  
,b.book_name book_namefrom borrow_record a  
INNER JOIN bookinfo b on b.id=a.book_id 
INNER JOIN userinfo c on c.uid=a.user_id 
where a.borrowtime=(select max(borrowtime)  
from borrow_record t where t.user_id=a.user_id) 
-- 说明：可以满足查询效果，不过性能不是最优解 

写法3 采用group by + join 性能最高，推荐采用

select a.user_id ,c.uname,a.borrowtime  
,b.book_name book_namefrom ( 
select t.user_id,max(borrowtime)  borrowtime 
from borrow_record t GROUP BY t.user_id) as e  
 
INNER JOIN  borrow_record a on e.user_id=a.user_id  
and e.borrowtime=a.borrowtimeINNER  
JOIN bookinfo b on b.id=a.book_id 
INNER JOIN userinfo c on c.uid=a.user_id 

运行效果如下：