SQL在日期范围内连续计数天数

Question

我正在计算一个人可能连续的天数，唯一的问题是我有日期范围，而不是直接的日期列表。这里有一个我所说的范围的例子：

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两列中的一个范围。

我试图实现的结果是：

Johnny has 9 consecutive days
Jenny  has 3 consecutive days

我遇到过一些解决方案的例子，但我找不到一个适合我的日期范围问题的解决方案。

迄今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

这是一个缺口和孤岛问题。一种选择是使用lag()和窗口sum()来构建相邻记录的组。然后，您可以按组进行聚合并计算连续天数，最后按名称筛选出最大连胜：

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

DB Fiddle上的演示：

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

Answer 2

这样的东西应该可以工作。用表格替换子选择。

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