MYSQL学习：GROUP BY分组取最新的一条记录

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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，希望对大家日常工作能有所帮助！

1、初始化数据表

-- 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_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
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_name
from (
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.borrowtime
INNER 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：

MYSQL学习：GROUP BY分组取最新的一条记录

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