Question

我到达一个月内使用服务的总天数。（Start_Date和End_Date都包括在内）

样本数据1：

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

Ans：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

Ans：服务使用天数= 6天。

如何编写代码来查找相同的代码，最好是SQL到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';

<强>输出：

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

<强>解释：

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

这将UNPIVOT表格，以便开始日期和结束日期在同一列（dt）中，并为开始提供相应的+1值，为结束日期提供-1

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。范围的开头始终为value=1和total=1，范围的结尾始终为total=0。如果日期在某个范围的中间，那么它将具有total>1或value=-1和total=1。使用此功能，如果您将value和total相乘，则范围的开头是value*total=1，范围的结尾是value*total=0，而任何其他值表示日期这是一个范围的中途。

这就是：

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start_end为NULL的日期，这会留下一个包含start和end行的表格，您可以使用LAG 1}}计算差异天数：

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

您需要做的只是SUM end - start的所有差异;它给出了上面的查询。

Answer 2

正如@Pravin Satav所说，你的要求并不是很清楚，我从你的解释中理解了这样的事情：

u.state_struct.loc0[:] = l

如何查找一个月内的总使用天数？

2 个答案: