SQL 加入：让双方的所有值都具有累积条件（Presto/AWS Athena）

Question

我研究这个看似简单的问题已经有一段时间了，但没有解决方案，假设我有一个带有日期列表的 table，另一个带有 phone 的 table数字、人物和日期，我需要一个包含所有姓名和所有日期的最终结果，第三列包含在与日期相同或大于日期的任何日期出现的唯一 phone 数字的数量结果，这是一个例子：

t1
+------------+
|    date    |
+------------+
| 01/01/2020 |
| 01/02/2020 |
| 01/03/2020 |
| 01/04/2020 |
| 01/05/2020 |
| 01/06/2020 |
| 01/07/2020 |
| 01/08/2020 |
+------------+

t2
+------+------------+--------------+
| name |    date    | phone_number |
+------+------------+--------------+
| John | 01/01/2020 |          123 |
| Mike | 01/02/2020 |          456 |
| Mike | 01/03/2020 |          789 |
| John | 01/04/2020 |          999 |
| Mike | 01/05/2020 |          111 |
| John | 01/06/2020 |          777 |
| Mike | 01/07/2020 |          123 |
| Mike | 01/08/2020 |          456 |
| John | 01/01/2020 |          789 |
| John | 01/02/2020 |          789 |
| Mike | 01/03/2020 |          789 |
| John | 01/04/2020 |          789 |
+------+------------+--------------+

我想要的结果：

+------+------------+-----------------------------------------------------------------+
| Name |   Month    | Comulative Unique Numbers (Unique Numbers in any date >= Month) |
+------+------------+-----------------------------------------------------------------+
| John | 01/01/2020 |                                                               4 |
| John | 01/02/2020 |                                                               3 |
| John | 01/03/2020 |                                                               3 |
| John | 01/04/2020 |                                                               3 |
| John | 01/05/2020 |                                                               1 |
| John | 01/06/2020 |                                                               1 |
| John | 01/07/2020 |                                                               0 |
| John | 01/08/2020 |                                                               0 |
| Mike | 01/01/2020 |                                                               4 |
| Mike | 01/02/2020 |                                                               4 |
| Mike | 01/03/2020 |                                                               4 |
| Mike | 01/04/2020 |                                                               3 |
| Mike | 01/05/2020 |                                                               3 |
| Mike | 01/06/2020 |                                                               2 |
| Mike | 01/07/2020 |                                                               2 |
| Mike | 01/08/2020 |                                                               1 |
+------+------------+-----------------------------------------------------------------+

我试了很多方法，这是我认为最接近的：

SELECT * FROM t1
LEFT OUTER JOIN
(SELECT t1.date, COUNT(DISTINCT phone_number) count, name FROM t1
LEFT OUTER JOIN
t2
ON t1.date < t2.date
GROUP BY t1.date,t2.name
ORDER BY 2 DESC) temp
ON t1.date = temp.date

最终结果中我仍然缺少行。

这就是我得到的：

+------+------------+-------+
| name |    date    | count |
+------+------------+-------+
| null | 2020-08-01 |     0 |
| John | 2020-01-01 |     3 |
| John | 2020-02-01 |     3 |
| John | 2020-03-01 |     3 |
| John | 2020-04-01 |     1 |
| John | 2020-05-01 |     1 |
| Mike | 2020-01-01 |     4 |
| Mike | 2020-02-01 |     4 |
| Mike | 2020-03-01 |     3 |
| Mike | 2020-04-01 |     3 |
| Mike | 2020-05-01 |     2 |
| Mike | 2020-06-01 |     2 |
| Mike | 2020-07-01 |     1 |
+------+------------+-------+

Answer 1

使用日历 table 方法，我们可以构建一个包含所有姓名和所有日期的参考 table。然后，将其加入包含实际数据的第二个 table：

SELECT
    b.name,
    a.date,
    COUNT(DISTINCT t.phone_number) AS unique_numbers
FROM t1 a
CROSS JOIN (SELECT DISTINCT name FROM t2) b
LEFT JOIN t2 t
    ON a.date = t.date AND b.name = t.name
GROUP BY
    b.name,
    a.date
ORDER BY
    b.name,
    a.date;

SQL 加入：让双方的所有值都具有累积条件（Presto/AWS Athena）

SQL Join: have all values from both sides with an accumulative condition (Presto/AWS Athena)

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