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。
最佳答案
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
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值 +1,结束日期指定为 -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。范围的开头总是
value=1
和 total=1
并且范围的结尾总是有 total=0
.如果日期在一个范围的中间,那么它要么是 total>1
或 value=-1
和 total=1
.使用这个,如果你乘以 value
和 total
那么范围的开始是 value*total=1
范围的结尾是 value*total=0
和任何其他值表示一个范围中途的日期。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
start_end
的日期。是 NULL
这将为您留下一张交替使用的 table start
和 end
您可以使用的行 LAG
计算天数差异: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
SUM
end - start
的所有差异;这给出了上面的查询。
关于sql - 如何找到一个月中使用的总天数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44705646/