SQL 计算日期范围内的连续天数

Question

我正在尝试计算一个人可能拥有的连续天数，唯一的问题是我有日期范围而不是直接的日期列表。这是我所说的范围的示例：

Name    Start_Date  End_Date
Johnny  2020-01-02  2020-01-04
Johnny  2020-01-05  2020-01-05
Johnny  2020-01-06  2020-01-10
Jenny   2020-02-07  2020-02-07
Jenny   2020-02-10  2020-02-11
Jenny   2020-02-12  2020-02-12

开始日期和结束日期是 2 列中的一个范围。

我想要达到的结果是这样的：

Johnny has 9 consecutive days
Jenny  has 3 consecutive days

我遇到过一些解决方案示例，但找不到适合我的日期范围问题的解决方案。

到目前为止使用的代码示例：

WITH
 
  dates(date, employee_number) AS (
    SELECT DISTINCT CAST(start_date AS DATE), name
    FROM myTABLE
    WHERE name = "Jenny"

  ),
   
  groups AS (
    SELECT
      ROW_NUMBER() OVER (ORDER BY date) AS rn, name,
      dateadd(day, -ROW_NUMBER() OVER (ORDER BY date), date) AS grp,
      date
    FROM dates
  )
SELECT
  name,
  COUNT(*) AS consecutiveDates,
  MIN(date) AS minDate,
  MAX(date) AS maxDate
FROM groups
GROUP BY grp, name

Answer 1

这是一个 gaps-and-islands 问题。一种选择是使用 lag() 和 window sum() 来构建相邻记录组。然后，您可以按组汇总并计算连续天数，最后按名称过滤最大的连胜：

select name, max(consecutive_days) consecutive_days
from (
    select name, datediff(day, min(start_date), max(end_date)) + 1 consecutive_days
    from (
        select t.*, 
            sum(case when start_date = dateadd(day, 1, lag_end_date) then 0 else 1 end) over(partition by name order by start_date) grp
        from (
            select t.*, 
                lag(end_date) over(partition by name order by start_date) lag_end_date
            from mytable t
        ) t
    ) t
    group by name, grp
) t
group by name

Demo on DB Fiddle:

name   | consecutive_days
:----- | ---------------:
Jenny  |                3
Johnny |                9

Answer 2

像这样的东西应该有用。将 sub-selects 替换为您的 table.

select name, DATEDIFF(dd, MIN(Start_date),MAX(end_date)) +1 from 

(
select a.name,a.start_date,b.end_date from 
(SELECT 'Johnny' name , '2020-01-02' start_date,  '2020-01-04' end_date
UNION SELECT 'Johnny' , '2020-01-05',  '2020-01-05'
UNION SELECT 'Johnny' , '2020-01-06',  '2020-01-10'
UNION SELECT 'Jenny'  , '2020-02-07',  '2020-02-07'
UNION SELECT 'Jenny'  , '2020-02-10',  '2020-02-11'
UNION SELECT 'Jenny'  , '2020-02-12',  '2020-02-12') a
LEFT join (SELECT 'Johnny' name , '2020-01-02' start_date,  '2020-01-04' end_date
UNION SELECT 'Johnny' , '2020-01-05',  '2020-01-05'
UNION SELECT 'Johnny' , '2020-01-06',  '2020-01-10'
UNION SELECT 'Jenny'  , '2020-02-07',  '2020-02-07'
UNION SELECT 'Jenny'  , '2020-02-10',  '2020-02-11'
UNION SELECT 'Jenny'  , '2020-02-12',  '2020-02-12') b on DATEADD(dd,1,a.end_date ) = b.start_date and a.name = b.name
)q

group by name

SQL 计算日期范围内的连续天数

SQL Counting Consecutive Days in Date Ranges

sql

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gaps-and-islands