确定总的周数,不包括重叠期。

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英文:

Determine total number of aggregate weeks excluding overlap periods

问题

我需要计算总共有多少个不重叠的周数,但要排除重叠的日期范围。我尝试过在表上进行自连接,但无法弄清楚。

所需输出:
期望的结果集将是一个新的结果集(具有连续的日期范围),排除日期重叠。例如,在上面的示例中,我想要删除从2018-09-08到2018-09-11的日期,因为这4天已经被上面的行覆盖。因此,它应该是根据时间顺序显示的所有行,具有合并的fromdate/todate。正如您所看到的,日期之间存在间隙,我不能使用Min(fromdate) | Max(todate)。

number name fromdate todate daycount
887 DAV 30/05/2018 29/05/2019 364
887 DAV 24/07/2019 07/08/2019 14
887 DAV 09/08/2019 10/08/2019 1
887 DAV 18/09/2019 17/11/2019 60
887 DAV 09/12/2019 17/02/2020 70
887 DAV 30/03/2020 30/03/2020 0
887 DAV 03/11/2021 02/03/2022 119
887 DAV 03/03/2022 30/03/2022 27
887 DAV 31/03/2022 05/07/2023 461

感谢您的帮助。

英文:

I have query which returns thousands of rows, with a number and name, fromdate and today. I need to calculate the total number of aggregate weeks but exclude the overlapping date range. I have not been able to find a way.

As you can see, the 7th row of data is 2018-09-12 and next starting date is 2018-09-08. All of these rows is only for one number, but in other number columns, there are a lot more overlaps but example is the same. I have to exclude these overlaps. e.g. in another example, row one is the whole month of January, and row two is a week from 10 to 17th in it. so this is another example.

I tried to do a self join on the table but could not figure it out.

select *
from #RW r1, #RW r2
where r1.number = r2.number and r1.name = r2.name 
and r1.Fromdate < r2.Fromdate and r1.todate > r2.Fromdate 
and r1.number = '887' 

**NOTE: If daycount = 0 , then that row should be ignored. it was generated based on another condition before I exported.

number name fromdate todate daycount
887 DAV NULL NULL 0
887 DAV 30/05/2018 05/06/2018 6
887 DAV 05/06/2018 19/06/2018 14
887 DAV 19/06/2018 18/07/2018 29
887 DAV 18/07/2018 15/08/2018 28
887 DAV 15/08/2018 12/09/2018 28
887 DAV 08/09/2018 17/10/2018 0
887 DAV 17/10/2018 07/11/2018 0
887 DAV 28/11/2018 09/01/2019 42
887 DAV 09/01/2019 05/02/2019 27
887 DAV 06/02/2019 29/05/2019 112
887 DAV 24/07/2019 07/08/2019 0
887 DAV 09/08/2019 09/08/2019 0
887 DAV 10/08/2019 10/08/2019 0
887 DAV 18/09/2019 17/11/2019 60
887 DAV 09/12/2019 20/01/2020 42
887 DAV 20/01/2020 17/02/2020 28
887 DAV 30/03/2020 30/03/2020 0
887 DAV 03/11/2021 10/11/2021 7
887 DAV 11/11/2021 14/12/2021 33
887 DAV 15/12/2021 31/01/2022 47
887 DAV 01/02/2022 27/02/2022 26
887 DAV 28/02/2022 02/03/2022 2
887 DAV 03/03/2022 30/03/2022 27
887 DAV 31/03/2022 12/04/2022 12
887 DAV 13/04/2022 27/04/2022 14
887 DAV 28/04/2022 24/05/2022 26
887 DAV 25/05/2022 22/06/2022 28
887 DAV 23/06/2022 20/07/2022 27
887 DAV 21/07/2022 16/08/2022 26
887 DAV 17/08/2022 14/09/2022 28
887 DAV 15/09/2022 09/11/2022 55
887 DAV 10/11/2022 07/12/2022 27
887 DAV 08/12/2022 11/01/2023 34
887 DAV 12/01/2023 05/02/2023 24
887 DAV 06/02/2023 21/02/2023 15
887 DAV 22/02/2023 23/03/2023 29
887 DAV 24/03/2023 08/05/2023 45
887 DAV 09/05/2023 05/06/2023 27
887 DAV 05/06/2023 05/07/2023 30

Desired output:
the desired result set would be to have a new result set (with consecutive date ranges) excluding the date overlaps. e.g. in above example, I want to remove the dates from 2018-09-08 until 2018-09-11 as these 4 days are already covered by the row above it. So naturally it should be all rows showing based on chronologica order, with merged fromdate/todate. as you can see, there are gaps between dates and I can't go with Min(fromdate) | Max(todate).

number name fromdate todate daycount
887 DAV 30/05/2018 29/05/2019 364
887 DAV 24/07/2019 07/08/2019 14
887 DAV 09/08/2019 10/08/2019 1
887 DAV 18/09/2019 17/11/2019 60
887 DAV 09/12/2019 17/02/2020 70
887 DAV 30/03/2020 30/03/2020 0
887 DAV 03/11/2021 02/03/2022 119
887 DAV 03/03/2022 30/03/2022 27
887 DAV 31/03/2022 05/07/2023 461

Thanks in advance for your help,

答案1

得分: 1

这是我的递归公共表达式(CTE)解决方案:

;WITH cteGeneratedDates AS (
    SELECT
        number,
        [name],
        fromdate,
        todate
    FROM TableName
    WHERE fromdate IS NOT NULL 
        AND daycount > 0 -- 如果应计算0天计数,请注释掉此行

    UNION ALL

    SELECT
        number,
        [name],
        fromdate = DATEADD(DAY, 1, fromdate),
        todate
    FROM cteGeneratedDates
    WHERE DATEADD(DAY, 1, fromdate) <= todate
),
cteGroupedConsecutiveDates AS (
    SELECT
        number,
        [name],
        fromdate,
        ROW_NUMBER() OVER (PARTITION BY number, [name] ORDER BY fromdate) todate
    FROM cteGeneratedDates
    GROUP BY number, [name], fromdate
)
SELECT 
    number,
    [name],
    MIN(fromdate) fromdate,
    MAX(fromdate) todate,
    DATEDIFF(DAY, MIN(fromdate), MAX(fromdate))
FROM cteGroupedConsecutiveDates
GROUP BY
    number,
    [name],
    DATEDIFF(DAY, fromdate, todate)
ORDER BY fromdate
OPTION (MAXRECURSION 2000)

尽管与您的预期输出有所不同,原因如下:为什么这些组没有在同一行下分组?

  • 03/11/2021 - 02/03/2022
  • 03/03/2022 - 30/03/2022
  • 31/03/2022 - 05/07/2023

在我的解决方案中,上述范围将在同一组中。

此外,07/11/2018和28/11/2018之间没有日期,您是否认为这个组需要拆分?30/05/2018 - 29/05/2019。

英文:

Here's my solution with recursive CTE

;WITH cteGeneratedDates AS (
        SELECT
			number,
			[name],
			fromdate,
			todate
		FROM TableName
		WHERE fromdate IS NOT NULL 
			AND daycount &gt; 0 -- comment this out if 0 daycount should be counted

        UNION ALL
		SELECT
			number,
			[name],
			fromdate = DATEADD(DAY, 1, fromdate),
			todate
        FROM cteGeneratedDates
		WHERE DATEADD(DAY, 1, fromdate) &lt;= todate
),
cteGroupedConsecutiveDates AS (
	SELECT
		number,
		[name],
		fromdate,
		ROW_NUMBER() OVER (PARTITION BY number, [name] ORDER BY fromdate) todate
	FROM cteGeneratedDates
	GROUP BY number, [name], fromdate
)
SELECT 
	number,
	[name],
	MIN(fromdate) fromdate,
	MAX(fromdate) todate,
	DATEDIFF(DAY, MIN(fromdate), MAX(fromdate))
FROM cteGroupedConsecutiveDates
GROUP BY
	number,
	[name],
	DATEDIFF(DAY, fromdate, todate)
ORDER BY fromdate
OPTION (MAXRECURSION 2000)

Though there's difference with your expected output because of the following: why do these groups not grouped under 1 row?

  • 03/11/2021 - 02/03/2022
  • 03/03/2022 - 30/03/2022
  • 31/03/2022 - 05/07/2023

In my solution, above ranges would be in the same group.

Also, there are no dates between 07/11/2018 and 28/11/2018, don't you think this group needs to be broken down? 30/05/2018 - 29/05/2019

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  • 本文由 发表于 2023年6月19日 09:03:49
  • 转载请务必保留本文链接:https://go.coder-hub.com/76503094.html
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