1074800250556052_个人页

胜哥 作者出的题目建表的sql能提供下吗？ 35题 select sum(count) as num,id from ( select count(*) count,requester_id as id from request_accepted GROUP BY requester_id UNION select count(*) count ,accepter_id as id from request_accepted GROUP BY accepter_id )t GROUP BY id ORDER BY num desc limit 1 33题 1.初始化: CREATE TABLE `Orders` ( `id` int(11) unsigned NOT NULL AUTO_INCREMENT, `shop_id` varchar(10) COLLATE utf8mb4_bin DEFAULT NULL, `pay_date` datetime DEFAULT NULL, `pay_no` int(11) DEFAULT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB AUTO_INCREMENT DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_bin INSERT INTO `Orders`(`id`, `shop_id`, `pay_date`, `pay_no`) VALUES (1, '1001', '2020-01-11 16:43:25', 5); INSERT INTO `Orders`(`id`, `shop_id`, `pay_date`, `pay_no`) VALUES (2, '1001', '2020-01-10 16:43:57', 3); INSERT INTO `Orders`(`id`, `shop_id`, `pay_date`, `pay_no`) VALUES (3, '1002', '2020-01-17 16:44:09', 4); INSERT INTO `Orders`(`id`, `shop_id`, `pay_date`, `pay_no`) VALUES (4, '1002', '2020-01-11 16:44:24', 2); INSERT INTO `Orders`(`id`, `shop_id`, `pay_date`, `pay_no`) VALUES (5, '1002', '2020-01-11 16:44:24', 1); INSERT INTO `Orders`(`id`, `shop_id`, `pay_date`, `pay_no`) VALUES (6, '1002', '2020-01-12 16:44:24', 0); 2.SQL 方法1：GROUP_CONCAT实现(数据库版本[5.7.16]) select * from Orders where id in ( select SUBSTRING_INDEX(GROUP_CONCAT(id ORDER BY pay_date,pay_no),',',1) from Orders GROUP BY shop_id ) 方法2：子查询实现(数据库版本[5.6.16]) select t.* from ( select * from Orders ORDER BY pay_date ,pay_no )t GROUP BY t.shop_id; 方法2有可能是不行的，不知道是数据库版本问题还是哪里设置没有 3.注意: GROUP_CONCAT有长度限制，记录过多需要注意设置以下配置 SET @@global.group_concat_max_len=数字