如何找到一个月中使用的总天数？

Question

我得到了一个月内使用服务的总天数。（Start_Date 和 End_Date 是 - 都包括在内）

示例数据 1：

User  Start_Date     End_Date
A     01-Jun-2017    30-Jun-2017
B     06-Jun-2017    30-Jun-2017

答：服务使用天数 = 30 天。

样本数据 2：

User  Start_Date     End_Date
C     06-Jun-2017    10-Jun-2017
D     02-Jun-2017    02-Jun-2017

答：服务使用天数 = 6 天。

如何编写代码以在 SQL 中找到相同的、更可取的 PLSQL。

Answer 1

测试数据：

CREATE TABLE your_table ( usr, start_date, end_date ) AS (
  SELECT 'A', DATE '2017-06-01', DATE '2017-06-03' FROM DUAL UNION ALL
  SELECT 'B', DATE '2017-06-02', DATE '2017-06-04' FROM DUAL UNION ALL -- Overlaps previous
  SELECT 'C', DATE '2017-06-06', DATE '2017-06-06' FROM DUAL UNION ALL
  SELECT 'D', DATE '2017-06-07', DATE '2017-06-07' FROM DUAL UNION ALL -- Adjacent to previous
  SELECT 'E', DATE '2017-06-11', DATE '2017-06-20' FROM DUAL UNION ALL
  SELECT 'F', DATE '2017-06-14', DATE '2017-06-15' FROM DUAL UNION ALL -- Within previous
  SELECT 'G', DATE '2017-06-22', DATE '2017-06-25' FROM DUAL UNION ALL
  SELECT 'H', DATE '2017-06-24', DATE '2017-06-28' FROM DUAL UNION ALL -- Overlaps previous and next
  SELECT 'I', DATE '2017-06-27', DATE '2017-06-30' FROM DUAL UNION ALL
  SELECT 'J', DATE '2017-06-27', DATE '2017-06-28' FROM DUAL;          -- Within H and I          

查询：

SELECT SUM( days ) AS total_days
FROM   (
  SELECT dt - LAG( dt ) OVER ( ORDER BY dt ) + 1 AS days,
         start_end
  FROM   (
    SELECT dt,
           CASE SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) * value
             WHEN 1 THEN 'start'
             WHEN 0 THEN 'end'
           END AS start_end
    FROM   your_table
    UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
  )
  WHERE start_end IS NOT NULL
)
WHERE start_end = 'end';

输出：

TOTAL_DAYS
----------
        25

说明：

SELECT dt, value
FROM   your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )

这将使UNPIVOT表中的开始日期和结束日期位于同一列 ( dt) 中，并为开始日期指定相应的 +1 值，为结束日期指定 -1 值。

SELECT dt,
       SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) AS total,
       value
FROM   your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )

将给出开始和结束日期以及这些生成值的累积总和。范围的开始总是有value=1和total=1范围的结束总是有total=0。如果日期在某个范围的中间，那么它将具有total>1orvalue=-1和total=1。利用这一点，如果你乘value和total随后一系列的开始是在value*total=1和一系列的到底是什么时候value*total=0和任何其它值都表示将通过一系列中途日期。

这就是它给出的：

SELECT dt,
       CASE SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) * value
         WHEN 1 THEN 'start'
         WHEN 0 THEN 'end'
       END AS start_end
FROM   your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )

然后，您可以过滤掉的日期时，start_end是NULL将离开你与交替的表start和end您可以使用行LAG来计算天差数：

SELECT dt - LAG( dt ) OVER ( ORDER BY dt ) + 1 AS days,
       start_end
FROM   (
  SELECT dt,
         CASE SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) * value
           WHEN 1 THEN 'start'
           WHEN 0 THEN 'end'
         END AS start_end
  FROM   your_table
  UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
)
WHERE start_end IS NOT NULL

您需要做的就是解决 ; 的SUM所有差异end - start。这给出了上面的查询。