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统计列日期范围内每个日期发生的行数

oracle9i plsql oracle sql

1

Will Dieterich · 技术社区 · 15 年前

我有一张表,上面有如下数据

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

我想提出一种每天都能计数的观点,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

我使用的是Oracle9,在如何处理这一问题上完全迷失了方向,我正在寻找任何让我开始工作的想法。
注意:生成一个表来保存日期是不实际的,因为我的最终产品必须分解到分钟。

3 回复 | 直到 15 年前

1

Cade Roux 15 年前

通常我用数字表来解决这类问题:

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

2

3

Quassnoi 15 年前

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

更新:

利用分析函数进行更好更快的查询。

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和在此日期之前结束的范围计数之间的差额。

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

此查询完成于 1 第二 1,000,000 排。

有关详细信息,请参阅我的日志:

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

3

2

Community Tales Farias 7 年前

您可以使用下面描述的方法,以便:

基本上:加入一个生成的日历并按列的子集分组。

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