LeetCode：Database 84.每次访问的交易次数

Posted 2021-06-27 Xiao Miao

要求：写一条 SQL 查询多少客户访问了银行但没有进行任何交易，多少客户访问了银行进行了一次交易等。

结果包含两列：
transactions_count：客户在一次访问中的交易次数
visits_count：在 transactions_count 交易次数下相应的一次访问时的客户数量
transactions_count 的值从 0 到所有用户一次访问中的 max(transactions_count) ，按 transactions_count 排序。

表: Visits的结构

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| visit_date    | date    |
+---------------+---------+
(user_id, visit_date) 是该表的主键
该表的每行表示 user_id 在 visit_date 访问了银行

表: Transactions的结构

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| user_id          | int     |
| transaction_date | date    |
| amount           | int     |
+------------------+---------+
该表没有主键，所以可能有重复行
该表的每一行表示 user_id 在 transaction_date 完成了一笔 amount 数额的交易
可以保证用户 (user) 在 transaction_date 访问了银行 (也就是说 Visits 表包含 (user_id, transaction_date) 行)

Visits 表:

+---------+------------+
| user_id | visit_date |
+---------+------------+
| 1       | 2020-01-01 |
| 2       | 2020-01-02 |
| 12      | 2020-01-01 |
| 19      | 2020-01-03 |
| 1       | 2020-01-02 |
| 2       | 2020-01-03 |
| 1       | 2020-01-04 |
| 7       | 2020-01-11 |
| 9       | 2020-01-25 |
| 8       | 2020-01-28 |
+---------+------------+

Transactions 表：

+---------+------------------+--------+
| user_id | transaction_date | amount |
+---------+------------------+--------+
| 1       | 2020-01-02       | 120    |
| 2       | 2020-01-03       | 22     |
| 7       | 2020-01-11       | 232    |
| 1       | 2020-01-04       | 7      |
| 9       | 2020-01-25       | 33     |
| 9       | 2020-01-25       | 66     |
| 8       | 2020-01-28       | 1      |
| 9       | 2020-01-25       | 99     |
+---------+------------------+--------+

Result Table：

+--------------------+--------------+
| transactions_count | visits_count |
+--------------------+--------------+
| 0                  | 4            |
| 1                  | 5            |
| 2                  | 0            |
| 3                  | 1            |
+--------------------+--------------+
* 对于 transactions_count = 0, visits 中 (1, "2020-01-01"), (2, "2020-01-02"), (12, "2020-01-01") 和 (19, "2020-01-03") 没有进行交易，所以 visits_count = 4 。
* 对于 transactions_count = 1, visits 中 (2, "2020-01-03"), (7, "2020-01-11"), (8, "2020-01-28"), (1, "2020-01-02") 和 (1, "2020-01-04") 进行了一次交易，所以 visits_count = 5 。
* 对于 transactions_count = 2, 没有客户访问银行进行了两次交易，所以 visits_count = 0 。
* 对于 transactions_count = 3, visits 中 (9, "2020-01-25") 进行了三次交易，所以 visits_count = 1 。
* 对于 transactions_count >= 4, 没有客户访问银行进行了超过3次交易，所以我们停止在 transactions_count = 3 。

SQL语句：

#1.方法1
with c as(
select a.user_id as u1,a.visit_date as v1,b.amount as a1
from visits a
left join transactions b
on a.user_id=b.user_id and a.visit_date=b.transaction_date
)

select e.r1 as transactions_count,count(c1)  as visits_count
from( 
select count(a1) as c1
from c
group by u1,v1)d 
right join (select row_number() over()-1 as r1
from c)e 
on e.r1=d.c1
where e.r1<=(select count(a1) as c1
from c
group by u1,v1
order by c1 desc
limit 1)
group by e.r1
order by e.r1 asc;
#2.方法2
with c as(
select count(b.amount)as c1
from visits a
left join transactions b
on a.user_id=b.user_id and a.visit_date=b.transaction_date
group by a.user_id,a.visit_date
)

select 
d.c1 as transactions_count,count(c.c1) as visits_count from
(select 0 as c1
union all
select row_number() over()  as c1
from transactions
)d
left join 
c 
on d.c1=c.c1 
where d.c1<=(select max(c1) from c)
group by d.c1
order by d.c1 asc