Question

我的表格中包含以下数据

Group       Start Date        End Date
A        01/01/01       01/03/01
A       01/01/01        01/02/01
A       01/03/01        01/04/01
B       01/01/01        01/01/01
ETC

我希望制作一个能够计算每一天的视图，就像这样

Group       Date        Count
A       01/01/01            2
A       01/02/01            2
A       01/03/01            2
A       01/04/01         1
B       01/01/01            1

我正在使用Oracle 9，我完全不知道如何处理这个问题，并且我正在寻找让我入门的任何想法。
注意：生成一个表来保存日期是不切实际的，因为我的最终产品必须分解为分钟。

Answer 1

WITH    q AS
        (
        SELECT  (
                SELECT  MIN(start_date)
                FROM    mytable
                ) + level - 1 AS mydate
        FROM    dual
        CONNECT BY
                level <= (
                SELECT  MAX(end_date) - MIN(start_date)
                FROM    mytable
                )
        )
SELECT  group, mydate,
        (
        SELECT  COUNT(*)
        FROM    mytable mi
        WHERE   mi.group = mo.group
                AND q BETWEEN mi.start_date AND mi.end_date
        ) 
FROM    q
CROSS JOIN
        (
        SELECT  DISTINCT group
        FROM    mytable
        ) mo

<强>更新

使用分析函数的更好，更快的查询。

主要思想是包含每个日期的范围数是在该日期之前开始的范围计数之前的差异以及在此之前结束的范围计数。

SELECT  cur_date,
        grouper,
        SUM(COALESCE(scnt, 0) - COALESCE(ecnt, 0)) OVER (PARTITION BY grouper ORDER BY cur_date) AS ranges
FROM    (
        SELECT  (
                SELECT  MIN(start_date)
                FROM    t_range
                ) + level - 1 AS cur_date
        FROM    dual
        CONNECT BY
                level <=
                (
                SELECT  MAX(end_date)
                FROM    t_range
                ) -
                (
                SELECT  MIN(start_date)
                FROM    t_range
                ) + 1
        ) dates
CROSS JOIN
        (
        SELECT  DISTINCT grouper AS grouper
        FROM    t_range
        ) groups
LEFT JOIN
        (
        SELECT  grouper AS sgrp, start_date, COUNT(*) AS scnt
        FROM    t_range
        GROUP BY
                grouper, start_date
        ) starts
ON      sgrp = grouper
        AND start_date = cur_date
LEFT JOIN
        (
        SELECT  grouper AS egrp, end_date, COUNT(*) AS ecnt
        FROM    t_range
        GROUP BY
                grouper, end_date
        ) ends
ON      egrp = grouper
        AND end_date = cur_date - 1
ORDER BY
        grouper, cur_date

此查询在1行的1,000,000秒内完成。

有关详细信息，请参阅我的博客中的此条目：

Oracle: generating a list of dates and counting ranges for each date

Answer 2

您可以使用这些SO中描述的方法：

基本上：使用生成的日历和GROUP BY列的子集连接。

SQL> WITH DATA AS (
  2  SELECT 'A' grp, to_date('01/01/01') start_date, to_date('01/03/01') end_date FROM DUAL
  3  UNION ALL SELECT 'A', to_date('01/01/01'), to_date('01/02/01') FROM DUAL
  4  UNION ALL SELECT 'A', to_date('01/03/01'), to_date('01/04/01') FROM DUAL
  5  UNION ALL SELECT 'B', to_date('01/01/01'), to_date('01/01/01') FROM DUAL
  6  ), calendar AS (
  7  SELECT to_date('01/01/01') + ROWNUM - 1 d
  8    FROM dual
  9    CONNECT BY LEVEL <= to_date('01/04/01') - to_date('01/01/01') + 1
 10  )
 11  SELECT data.grp, calendar.d, COUNT(*) cnt
 12    FROM data
 13    JOIN calendar ON calendar.d BETWEEN data.start_date AND data.end_date
 14   GROUP BY data.grp, calendar.d;

GRP D                  CNT
--- ----------- ----------
A   04/01/2001           1
A   02/01/2001           2
B   01/01/2001           1
A   03/01/2001           2
A   01/01/2001           2

Answer 3

通常我用数字表解决这类问题：

WITH Dates AS (
    SELECT DateAdd(d, Numbers.Number - 1, '1/1/2001') AS Date
    FROM Numbers
    WHERE Numbers.Number BETWEEN 1 AND 100000 -- Arbitrary date range
)
SELECT GroupData.Group, Dates.Date, COUNT(*)
FROM Dates
LEFT JOIN GroupData
    ON Dates.Date BETWEEN GroupData.StartDate AND GroupData.EndDate
GROUP BY GroupData.Group, Dates.Date
ORDER BY GroupData.Group, Dates.Date

计算列日期范围中每个日期出现的行数

3 个答案: