MySQL 时间范围，不包括重叠的时间范围

Question

id	start_date	end_date
7	2021-08-07 15:25:10.000000	2021-08-10 15:25:10.000000
8	2021-08-09 15:25:10.000000	2021-08-12 15:25:10.000000
9	2021-08-19 15:25:10.000000	2021-08-22 15:25:10.000000
10	2021-08-21 15:25:10.000000	2021-08-25 15:25:10.000000

我有这样的时间数据重叠。我要做的是取开始和结束而不计算中间的那些，然后计算它们之间的差距。所以我期望的结果是

start_date	end_date
2021-08-07 15:25:10.000000	2021-08-12 15:25:10.000000
2021-08-19 15:25:10.000000	2021-08-25 15:25:10.000000

我这样写了一个简单的查询，但是结果和差错了，我该怎么办？ （我的表只包含 id、start_date 和 end_date）

SELECT *,TIME_TO_SEC(TIMEDIFF(end_date, start_date)) / 3600 AS Total
FROM (
  SELECT   DISTINCT *
        ,(SELECT MIN(start_date) FROM deneme
          WHERE end_date BETWEEN t.start_date AND t.end_date) AS TIME_ENTER
        ,(SELECT MAX(end_date) FROM deneme
          WHERE start_date BETWEEN t.start_date AND t.end_date) AS TIME_EXIT
  FROM deneme t
  ) AS o
GROUP BY id;

Answer 1

一个想法：

列出没有记录的最小和最大日期，然后用它来区分之前和之后（无记录日期范围）作为一个单独的组。

*完整的分解，请参考这个小提琴链接：https://dbfiddle.uk/?rdbms=mariadb_10.4&fiddle=d366ff58efc1f520132ed7cd81d729de

细分：

1. 根据表中的现有数据生成日期范围。幸运的是，MariaDB 10.4 支持WITH RECURSIVE 的窗口功能，因此一旦您了解了WITH RECURSIVE 的工作原理，就很容易生成日期范围：

WITH RECURSIVE cte AS (
   SELECT MIN(start_date) AS dt, MAX(end_date) AS mxdt FROM deneme UNION ALL
    SELECT dt+INTERVAL 1 DAY, mxdt FROM cte WHERE dt+INTERVAL 1 DAY <= mxdt)

SELECT * FROM cte;

1. 将生成的日期范围和LEFT JOIN与表deneme一起使用，只获取没有记录的日期，获取MIN()和MAX()的日期：

WITH RECURSIVE cte AS (
   SELECT MIN(start_date) AS dt, MAX(end_date) AS mxdt FROM deneme UNION ALL
    SELECT dt+INTERVAL 1 DAY, mxdt FROM cte WHERE dt+INTERVAL 1 DAY <= mxdt)
    
SELECT MIN(dt) min_no_record, MAX(dt) max_no_record
FROM cte 
LEFT JOIN deneme d1 ON dt BETWEEN d1.start_date AND d1.end_date
WHERE d1.start_date IS NULL;

1. 将上面的查询设为子查询并再次将CROSS JOIN 与表deneme 一起使用，但这次使用CASE 表达式来分配在不存在的记录期之前和之后存在的每个日期：

WITH RECURSIVE cte AS (
   SELECT MIN(start_date) AS dt, MAX(end_date) AS mxdt FROM deneme UNION ALL
    SELECT dt+INTERVAL 1 DAY, mxdt FROM cte WHERE dt+INTERVAL 1 DAY <= mxdt)
    
SELECT *,
       CASE WHEN d.end_date < min_no_record THEN min_no_record
            WHEN d.start_date > max_no_record THEN max_no_record
            ELSE 0 END
FROM deneme d 
CROSS JOIN   
(SELECT MIN(dt) min_no_record, MAX(dt) max_no_record
FROM cte 
LEFT JOIN deneme d1 ON dt BETWEEN d1.start_date AND d1.end_date
WHERE d1.start_date IS NULL) v;

*CASE 表达式中... THEN min_no_record 或... THEN max_no_record 的部分可以是任何东西。您可以将其定义为... THEN 1 或... THEN 2，但这取决于您。

4. 最后，获取组的MIN() 和MAX() 日期。上面的CASE 表达式在GROUP BY 中使用；从SELECT 中省略：

WITH RECURSIVE cte AS (
    SELECT MIN(start_date) AS dt, MAX(end_date) AS mxdt FROM deneme UNION ALL
    SELECT dt+INTERVAL 1 DAY, mxdt FROM cte WHERE dt+INTERVAL 1 DAY <= mxdt)
 SELECT MIN(d.start_date),
        MAX(d.end_date)
   FROM deneme d 
    CROSS JOIN
  (SELECT MIN(dt) min_no_record, MAX(dt) max_no_record
     FROM cte 
     LEFT JOIN deneme d1 ON dt BETWEEN d1.start_date AND d1.end_date
   WHERE d1.start_date IS NULL) v
GROUP BY CASE WHEN d.end_date < min_no_record THEN min_no_record
            WHEN d.start_date > max_no_record THEN max_no_record
            ELSE 0 END;