如何将开始日期和结束日期与其他开始日期和结束日期分开？

Question

我在下面编写了两个查询，结果类似于：

表：company_changes

user_id	start_at	end_at	company_id
189	2020-12-12	2021-03-02	88
189	2021-03-02	2050-01-01	169

表：enablement_changes

user_id	start_at	end_at	enablement
189	2020-12-12	2021-10-15	disabled
189	2021-10-15	2050-01-01	enabled

重要的是我知道用户何时处于某个company_id 并且是enabled 或disabled。

我想要的结果是这样的表格：

user_id	start_at	end_at	company_id	status
189	2020-12-12	2021-03-02	88	disabled
189	2021-03-02	2021-10-15	169	disabled
189	2021-10-15	2050-01-01	169	enabled

我本质上想将这些查询的结果组合在一起。 2050-01-01 是未来的任意日期。由于user_id 没有更改status 或company_id，因此它显示为2050-01-01，因为它是用户的当前状态。

知道如何解决这个问题吗？

这里是小提琴：http://sqlfiddle.com/#!9/5c42b6

第一次在 Stackoverflow 上提问...如果我的问题格式不正确，请告诉我。

Answer 1

如果在实践中您有更复杂的数据并且可能存在重叠的时间间隔，例如：

表：enablement_changes

user_id	start_at	end_at	enablement
189	2020-12-12	2021-10-15	disabled
189	2020-12-20	2021-02-10	enabled
189	2021-10-15	2050-01-01	enabled

我推荐一个更复杂的解决方案：

WITH _k AS (
    SELECT 1 AS n
    UNION ALL
    SELECT 2 AS n
), _points AS (
  SELECT user_id, CASE WHEN n = 1 THEN start_at ELSE end_at END AS date_point, n
    FROM company_changes
   CROSS JOIN _k
   UNION
  SELECT user_id, CASE WHEN n = 1 THEN start_at ELSE end_at END AS date_point, n
    FROM enablement_changes
   CROSS JOIN _k
), _drank AS (
  SELECT p.user_id, p.date_point, DENSE_RANK() OVER(PARTITION BY p.user_id ORDER BY p.date_point) AS dr
    FROM _points AS p
   GROUP BY p.user_id, p.date_point
)
SELECT d1.user_id, d1.date_point AS start_at, d2.date_point AS end_at, c.company_id, MAX(s.status) AS status -- or MIN if status disabled is stronger than enabled in the same time
  FROM _drank AS d1
  JOIN _drank AS d2 ON d1.dr = d2.dr-1 AND d1.user_id = d2.user_id
  LEFT JOIN company_changes AS c    ON d1.user_id = c.user_id AND d1.date_point < c.end_at AND c.start_at < d2.date_point 
  LEFT JOIN enablement_changes AS s ON d1.user_id = s.user_id AND d1.date_point < s.end_at AND s.start_at < d2.date_point 
 GROUP BY d1.user_id, d1.date_point, d2.date_point, c.company_id
 ORDER BY 1,2,3;

db<>fiddle demo

输出：

Answer 2

Lukasz 解决方案很好。

but 将匹配c 表结束时间与e 表开始时间匹配的行。通常日期时间范围希望包含 start 但不匹配 end 否则您将得到两行。

它会错过任何在c 开始和结束于e 表行之后的连接，但您想要匹配子集。后一点取决于您是进行密集匹配（始终有行）还是稀疏匹配（有时只有行）

第一个问题可以通过额外的检查来解决：

SELECT c.user_id 
       ,GREATEST(c.start_at, e.start_at) AS start_at
       ,LEAST(c.end_at, e.end_at) AS end_at
       ,c.company_id 
       ,e.status
FROM company_changes c
JOIN enablement_changes e
  ON (c.start_at BETWEEN e.start_at AND e.end_at AND c.start_at < e.end_at
    OR c.end_at BETWEEN e.start_at AND e.end_at AND c.end_at >  e.start_at )
    AND c.user_id = e.user_id
ORDER BY 1,2;

在哪里匹配你需要的稀疏匹配。

SELECT c.user_id 
       ,GREATEST(c.start_at, e.start_at) AS start_at
       ,LEAST(c.end_at, e.end_at) AS end_at
       ,c.company_id 
       ,e.status
FROM company_changes c
JOIN enablement_changes e
    ON c.user_id = e.user_id
        AND (c.end_at > e.start AND c.start_at < e.end_at)
ORDER BY 1,2;

在具有大范围的非常大的表上，后面的代码可能很昂贵

Answer 3

使用JOIN 和BETWEEN：

SELECT c.user_id 
       ,GREATEST(c.start_at, e.start_at) AS start_at
       ,LEAST(c.end_at, e.end_at) AS end_at
       ,c.company_id 
       ,e.status
FROM company_changes c
JOIN enablement_changes e
  ON (c.start_at BETWEEN e.start_at AND e.end_at
    OR c.end_at BETWEEN e.start_at AND e.end_at)
    AND c.user_id = e.user_id
ORDER BY 1,2;

db<>fiddle demo

输出：