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查询单日多次下订单的用户信息?

[编程题]查询单日多次下订单的用户信息?
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现有某公司部分订单数据及用户会员等级数据,如下所示:
订单信息表:order_tb(订单id-order_id,用户id-user_id,订单金额-order_price,订单创建时间-order_time)
order_id user_id order_price order_time
101 11 380 2022-09-01 09:00:00
102 12 200 2022-09-01 10:00:00
103 13 260 2022-09-01 12:00:00
104 11 100 2022-09-02 11:00:00
105 12 150 2022-09-02 12:00:00
106 12 1200 2022-09-02 13:00:00
107 11 60 2022-09-03 09:00:00
108 13 380 2022-09-03 09:30:00
会员等级信息表:uservip_tb用户id-user_id,会员等级-vip,积分-point
user_id vip point
10 银卡会员 530
11 银卡会员 1555
12 钻石会员 12000
13 金卡会员 6115
14 普通会员 230
15 银卡会员 810
16 普通会员 330
请查询单日下单多次的用户信息?
要求输出:订单日期,user_id,下单次数,会员等级
注:单日多次下订单指该日同一用户下单次数大于1次,结果按照下单次数降序排序
示例数据结果如下:
order_date user_id order_nums vip
2022-09-02 12 2 钻石会员

结果解释:
user_id为12的用户在9月2日分别下了order_id为105、106的订单,故结果如上。
示例1

输入

drop table if exists  `order_tb` ; 
CREATE TABLE `order_tb` (
`order_id` int(11) NOT NULL,
`user_id` int(11) NOT NULL,
`order_price` int(11) NOT NULL,
`order_time` datetime NOT NULL,
PRIMARY KEY (`order_id`));
INSERT INTO order_tb VALUES(101,11,380,'2022-09-01 09:00:00'); 
INSERT INTO order_tb VALUES(102,12,200,'2022-09-01 10:00:00'); 
INSERT INTO order_tb VALUES(103,13,260,'2022-09-01 12:00:00'); 
INSERT INTO order_tb VALUES(104,11,100,'2022-09-02 11:00:00'); 
INSERT INTO order_tb VALUES(105,12,150,'2022-09-02 12:00:00'); 
INSERT INTO order_tb VALUES(106,12,1200,'2022-09-02 13:00:00'); 
INSERT INTO order_tb VALUES(107,11,60,'2022-09-03 09:00:00'); 
INSERT INTO order_tb VALUES(108,13,380,'2022-09-03 09:30:00'); 

drop table if exists  `uservip_tb` ; 
CREATE TABLE `uservip_tb` (
`user_id` int(11) NOT NULL,
`vip` varchar(16) NOT NULL,
`point` int(11) NOT NULL,
PRIMARY KEY (`user_id`));
INSERT INTO uservip_tb VALUES(10,'银卡会员',530); 
INSERT INTO uservip_tb VALUES(11,'银卡会员',1555); 
INSERT INTO uservip_tb VALUES(12,'钻石会员',12000); 
INSERT INTO uservip_tb VALUES(13,'金卡会员',6115); 
INSERT INTO uservip_tb VALUES(14,'普通会员',230); 
INSERT INTO uservip_tb VALUES(15,'银卡会员',810); 
INSERT INTO uservip_tb VALUES(16,'普通会员',330);

输出

order_date|user_id|order_nums|vip
2022-09-02|12|2|钻石会员
select substring_index(order_time," ",1) as order_date,
       t1.user_id,
       count(*) as order_nums,
       vip
from order_tb t1 left outer join uservip_tb t2 on t1.user_id=t2.user_id
group by order_date,t1.user_id
having count(*)>1;

select a1.da order_date,a1.user_id,a1.cn order_nums,a1.vip
from
    (select distinct t1.user_id,
       substring_index(order_time," ",1) as da,
       vip,
       count(*) over(partition by substring_index(order_time," ",1),t1.user_id) as cn
    from order_tb t1
    left outer join uservip_tb t2
    on t1.user_id=t2.user_id) a1
    where a1.cn>1;

substring_index可以替换成其他提取日期的函数,包括date()、date_format()等
发表于 2025-06-25 14:16:01 回复(0)