
SQL64 对顾客ID和日期排序
SQL64 对顾客ID和日期排序
描述
有Orders表
cust_id | order_num | order_date |
---|---|---|
andy | aaaa | 2021-01-01 00:00:00 |
andy | bbbb | 2021-01-01 12:00:00 |
bob | cccc | 2021-01-10 12:00:00 |
dick | dddd | 2021-01-11 00:00:00 |
【问题】编写 SQL 语句,从 Orders 表中检索顾客 ID(cust_id)和订单号(order_num),并先按顾客 ID 对结果进行排序,再按订单日期倒序排列。
【示例答案】
返回2列,cust_id和order_num
cust_id | order_num |
---|---|
andy | bbbb |
andy | aaaa |
bob | cccc |
dick | dddd |
【示例解析】
首先根据cust_id进行排列,andy在bob和dick前,再根据order_date进行排列,订单号bbbb的订单时间是 "2021-01-01 12:00:00"大于订单号aaaa的时间"2021-01-01 00:00:00"
示例1
输入:
DROP TABLE IF EXISTS `Orders`;
CREATE TABLE IF NOT EXISTS `Orders` (
`cust_id` varchar(255) NOT NULL COMMENT '顾客 ID',
`order_num` varchar(255) NOT NULL COMMENT '订单号',
`order_date` timestamp NOT NULL COMMENT '订单时间'
);
INSERT INTO `Orders` VALUES ('andy','aaaa','2021-01-01 00:00:00'),
('andy','bbbb','2021-01-01 12:00:00'),
('bob','cccc','2021-01-10 12:00:00'),
('dick','dddd','2021-01-11 00:00:00');
复制
输出:
andy|bbbb
andy|aaaa
bob|cccc
dick|dddd
我的答案
select cust_id,order_num from Orders order by cust_id asc,order_date desc
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