SQL面试题（2）

时间：2024-03-07 08:11:21

 第一题

创建trade_orders表：
create table `trade_orders`(
`trade_id` varchar(255) NULL DEFAULT NULL,
`uers_id` varchar(255),
`trade_fee` int(20),
`product_id` varchar(255),
`time` varchar(255)
)ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_0900_ai_ci ROW_FORMAT = dynamic
添加数据：
INSERT INTO `trade_orders` VALUES ('20171023140109701120.0', 'ovsuXw6-7XI81U049TbjugEz912w', '168', '2017001000030.0', '2017-10-23 22:04:45');
INSERT INTO `trade_orders` VALUES ('20171023220445700096.0', 'ovsuXw2TajDGuHPSF798z9FSy38c', '18', '2017001000031.0', '2017-10-23 11:55:23');
INSERT INTO `trade_orders` VALUES ('20171023110103801856.0', 'ovsuXw6-7XI81U049TbjugEz912w', '', '2017002000130.0', '2017-10-28 21:19:18');
INSERT INTO `trade_orders` VALUES ('20171023184746299392.0', 'ovsuXw6-7XI81U049TbjugEz912w', '19', '2017002000129.0', '2017-10-28 21:19:18');
INSERT INTO `trade_orders` VALUES ('20171023184832901120.0', 'ovsuXw 2114bvpcjiR83R4dVzPw2Gg', '200', '2017002000129.0', '2023-03-30 11:44:07');
INSERT INTO `trade_orders` VALUES ('20171023205014200320.0', 'ovsuXwwqmoj-YC987zDnpE9KaPs8', '300', '2017001000031.0', '2023-03-29 18:48:32');
INSERT INTO `trade_orders` VALUES ('20171023110103801856.0', 'ovsuXw6-7XI81U049TbjugEz912w', '0', '2017002000130.0', '2017-10-28 21:19:18');

题一：每个用户的首次付费时间

select
	uers_id as 用户id,
	min(time) as 首次付费时间
from 
	trade_orders
GROUP BY
	uers_id

 题二：每个用户首次付费后30天内的总付费次数：

SELECT 
	users_id AS 用户ID, 
	COUNT(distinct trade_fee) AS 总付费次数 
FROM 
	trade_orders AS t
WHERE 
	time <= DATE_ADD(
	(
	SELECT 
		MIN(time) 
	FROM 
		trade_orders AS min_time 
	WHERE 
		min_time.users_id = t.users_id and t.trade_fee != ""), INTERVAL 30 DAY
	) 
GROUP BY 
	users_id
ORDER BY
	总付费次数 desc;

 解释：

  1. 首先，从交易订单表（trade_orders）中选择用户ID（users_id）和不同交易费用（trade_fee）的数量&

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