SQL

Question

我有一张这样的桌子：

如果 t1_view_dt 彼此相隔 14 天以内，我需要在名为 t2_view_dt 的另一列中显示每组的第一行。

+------------+------+------------+------------+--------+
| row number | user | t1_view_dt | t2_view_dt | series |
+------------+------+------------+------------+--------+
|          1 |  123 |        1/1 |        1/1 | 1      |
+------------+------+------------+------------+--------+
|          2 |  123 |       1/10 |       1/10 | 1      |
+------------+------+------------+------------+--------+
|          3 |  123 |       1/20 |       1/20 | 2      |
+------------+------+------------+------------+--------+
|          4 |  123 |       1/25 |       1/20 | 2      |
+------------+------+------------+------------+--------+
|          5 |  123 |       1/30 |       1/20 | 2      |
+------------+------+------------+------------+--------+
|          6 |  123 |       2/10 |       2/10 | 3      |
+------------+------+------------+------------+--------+
|          7 |  123 |       2/15 |       2/10 | 3      |
+------------+------+------------+------------+--------+

例如，第 1 行和第 2 行属于同一组，因为组开始日期是 1/1，而 1/10 在从 1/1（第 1 行）开始的 14 天 window 内。日期为 1/20 的第 3 行在第 14 天之外 window，因此属于第 2 组（系列 2）等

我该怎么做？如何按 14 天划分日期以及如何根据此规则从一个新组开始？

Answer 1

您还没有提供 RDBMS。
这是 mySQL / mariaDB 的解决方案。

create table t1 (
user int,
view_dt date);

insert into t1 values
(122,'2022-01-02'),
(123,'2022-01-01'),
(123,'2022-01-10'),
(123,'2022-01-20'),
(123,'2022-01-25'),
(123,'2022-01-30'),
(123,'2022-02-10'),
(123,'2022-02-15');

SET @gNumber = 0;
SET @gDate = '1900-01-01';
set @user = 1;
SELECT
  @gNumber := case 
    when @user = user and datediff(view_dt,@gDate) <= 14 then @gNumber 
    else @gNumber + 1 end groupNumber,
  @user := user user,
  view_dt,
  @gDate := case when datediff(view_dt,@gDate) <= 14 then @gDate
    else view_dt end groupDate
FROM t1 
ORDER BY 
  user,
  view_dt;

groupNumber | user | view_dt    | groupDate 
----------: | ---: | :--------- | :---------
          1 |  122 | 2022-01-02 | 2022-01-02
          2 |  123 | 2022-01-01 | 2022-01-02
          2 |  123 | 2022-01-10 | 2022-01-02
          3 |  123 | 2022-01-20 | 2022-01-20
          3 |  123 | 2022-01-25 | 2022-01-20
          3 |  123 | 2022-01-30 | 2022-01-20
          4 |  123 | 2022-02-10 | 2022-02-10
          4 |  123 | 2022-02-15 | 2022-02-10

db<>fiddle here

SQL - SQL Server 2019：将间隔小于 14 天的行分组，并且 return 每组的第 1 行

SQL - SQL Server 2019: Group rows with that are less than 14 days apart and return the 1 row of each group

partition