题目
表: Visits
+---------------+---------+
| 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
+------------------+---------+
| 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)
银行想要得到银行客户在一次访问时的交易次数和相应的在一次访问时该交易次数的客户数量的图表
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
下面是查询结果格式的例子：
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 |
+---------+------------------+--------+
结果表:
+--------------------+--------------+
| 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 。
创建数据
CREATE TABLE Visits(
user_id INT,
visit_date DATE);
CREATE TABLE Transactions1(
user_id INT,
transaction_date DATE,
amount INT);
INSERT INTO Visits VALUE(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');
INSERT INTO Transactions1 VALUE(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);
解答
先统计每个id 每个时间的交易数量
SELECT T.`user_id`, T.`transaction_date`, COUNT(T.`amount`)
FROM Transactions1 AS T
GROUP BY T.`user_id`, T.`transaction_date`；
两表连接
SELECT *
FROM Visits AS V
LEFT JOIN (SELECT T.`user_id`, T.`transaction_date`, COUNT(T.`amount`) AS cnt
FROM Transactions1 AS T
GROUP BY T.`user_id`, T.`transaction_date`) AS tmp
ON V.`user_id` = tmp.user_id AND V.`visit_date` = tmp.`transaction_date`;
然后对cnt为null的转为0然后统计数量即可
SELECT IFNULL(tmp.cnt, 0) AS transactions_count, COUNT(IFNULL(tmp.cnt, 0)) AS visits_count
FROM Visits AS V
LEFT JOIN (SELECT T.`user_id`, T.`transaction_date`, COUNT(T.`amount`) AS cnt
FROM Transactions1 AS T
GROUP BY T.`user_id`, T.`transaction_date`) AS tmp
ON V.`user_id` = tmp.user_id AND V.`visit_date` = tmp.`transaction_date`
GROUP BY transactions_count;
但是这里没有2
需要生成一张从0到上面这张表的最大值(设为N)的表和它进行JOIN，来获得最终结果。
考虑到最大出现的数字，肯定不会超过Transaction表的行数，使用Transaction表：
SELECT
(@t := @t + 1) AS id
FROM
Transactions1 t0,
(SELECT
@t := - 1) init ;
选出每个用户每天交易次数的最大值
SELECT MAX(cnt) AS max_cnt
FROM (SELECT T.`user_id`, T.`transaction_date`, COUNT(T.`amount`) AS cnt
FROM Transactions1 AS T
GROUP BY T.`user_id`, T.`transaction_date`) AS tmp;
取出以上序列小于等于交易次数最大值的部分即可
SELECT all_c.id
FROM
(SELECT
(@t := @t + 1) AS id
FROM
Transactions1 t0,
(SELECT
@t := - 1) init) AS all_c,
(SELECT
MAX(cnt) AS max_cnt
FROM
(SELECT
T.`user_id`,
T.`transaction_date`,
COUNT(T.`amount`) AS cnt
FROM
Transactions1 AS T
GROUP BY T.`user_id`,
T.`transaction_date`) AS tmp) AS max_c
WHERE all_c.id <=max_c.max_cnt;
与上边的结果进行左连接即可
SELECT A.id AS transactions_count, IFNULL(B.visits_count, 0)
FROM (SELECT all_c.id
FROM
(SELECT
(@t := @t + 1) AS id
FROM
Transactions1 t0,
(SELECT
@t := - 1) init) AS all_c,
(SELECT
MAX(cnt) AS max_cnt
FROM
(SELECT
T.`user_id`,
T.`transaction_date`,
COUNT(T.`amount`) AS cnt
FROM
Transactions1 AS T
GROUP BY T.`user_id`,
T.`transaction_date`) AS tmp) AS max_c
WHERE all_c.id <=max_c.max_cnt) AS A
LEFT JOIN (SELECT IFNULL(tmp.cnt, 0) AS transactions_count, COUNT(IFNULL(tmp.cnt, 0)) AS visits_count
FROM Visits AS V
LEFT JOIN (SELECT T.`user_id`, T.`transaction_date`, COUNT(T.`amount`) AS cnt
FROM Transactions1 AS T
GROUP BY T.`user_id`, T.`transaction_date`) AS tmp
ON V.`user_id` = tmp.user_id AND V.`visit_date` = tmp.`transaction_date`
GROUP BY transactions_count) AS B
ON A.id = B.transactions_count
貌似是写过最长sql代码了