Sql 找到一个';运行';排序结果集中的行数
我试图找出一种方法来识别满足某些条件的结果(按顺序连续行)的“运行”。目前,我正在订购一个结果集,并用眼睛扫描特定的模式。下面是一个例子:Sql 找到一个';运行';排序结果集中的行数,sql,oracle,gaps-and-islands,Sql,Oracle,Gaps And Islands,我试图找出一种方法来识别满足某些条件的结果(按顺序连续行)的“运行”。目前,我正在订购一个结果集,并用眼睛扫描特定的模式。下面是一个例子: SELECT the_date, name FROM orders WHERE the_date BETWEEN to_date('2013-09-18',..) AND to_date('2013-09-22', ..) ORDER BY the_date --------------------------
SELECT the_date, name
FROM orders
WHERE
the_date BETWEEN
to_date('2013-09-18',..) AND
to_date('2013-09-22', ..)
ORDER BY the_date
--------------------------------------
the_date | name
--------------------------------------
2013-09-18 00:00:01 | John
--------------------------------------
2013-09-19 00:00:01 | James
--------------------------------------
2013-09-20 00:00:01 | John
--------------------------------------
2013-09-20 00:00:02 | John
--------------------------------------
2013-09-20 00:00:03 | John
--------------------------------------
2013-09-20 00:00:04 | John
--------------------------------------
2013-09-21 16:00:01 | Jennifer
--------------------------------------
我想从这个结果集中提取的是2013-09-20
上归属于John
的所有行。一般来说,我要查找的是来自同一名称的运行结果,一行>=3。我正在使用Oracle 11,但我想知道这是否可以通过严格的SQL实现,或者是否必须使用某种分析功能。试试这个
WITH ORDERS
AS (SELECT
TO_DATE ( '2013-09-18 00:00:01',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'John' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-19 00:00:01',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'James' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-20 00:00:01',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'John' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-20 00:00:02',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'John' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-20 00:00:03',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'John' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-20 00:00:04',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'John' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-21 16:00:01',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'Jennifer' AS NAME
FROM
DUAL)
SELECT
B.*
FROM
(SELECT
TRUNC ( THE_DATE ) THE_DATE,
NAME,
COUNT ( * )
FROM
ORDERS
WHERE
THE_DATE BETWEEN TRUNC ( TO_DATE ( '2013-09-18',
'YYYY-MM-DD' ) )
AND TRUNC ( TO_DATE ( '2013-09-22',
'YYYY-MM-DD' ) )
GROUP BY
TRUNC ( THE_DATE ),
NAME
HAVING
COUNT ( * ) >= 3) A,
ORDERS B
WHERE
A.NAME = B.NAME
AND TRUNC ( A.THE_DATE ) = TRUNC ( B.THE_DATE );
输出
9/20/2013 12:00:01 AM John
9/20/2013 12:00:02 AM John
9/20/2013 12:00:03 AM John
9/20/2013 12:00:04 AM John
您需要多个嵌套窗口函数:
SELECT *
FROM
(
SELECT the_date, name, grp,
COUNT(*) OVER (PARTITION BY grp) AS cnt
FROM
(
SELECT the_date, name,
SUM(flag) OVER (ORDER BY the_date) AS grp
FROM
(
SELECT the_date, name,
CASE WHEN LAG(name) OVER (ORDER BY the_date) = name THEN 0 ELSE 1 END AS flag
FROM orders
WHERE
the_date BETWEEN
TO_DATE('2013-09-18',..) AND
TO_DATE('2013-09-22', ..)
) dt
) dt
) dt
WHERE cnt >= 3
ORDER BY the_date
你能把预期的结果公布吗?你所说的运行是什么意思?@realspirituals我解释了预期的输出以及我所说的“运行”结果是什么意思。这还不清楚吗?请看我的帖子,并确认这就是您要寻找的……对不起,我在定义“运行”时没有正确解释自己,而且我没有提供足够好的数据示例。我要找的是,当同一个人按顺序出现3行或更多行时。如果Jane在同一天有4行,但它们不符合顺序,则不应返回她的数据。不过我会更新这个问题。真的很酷的解决方案。不知道滞后函数。在接下来的几天里,我也会花一些时间真正熟悉。