2

I'm trying to figure out a way of identifying a "run" of results (successive rows, in order) that meet some condition. Currently, I'm ordering a result set, and scanning by eye for particular patterns. Here's an example:

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
--------------------------------------

What I want to extract from this result set is all the rows attributed to John on 2013-09-20. Generally what I'm looking for is a run of results from the same name, in a row, >= 3. I'm using Oracle 11, but I'm interested to know if this can be achieved with strict SQL, or if some kind of analytical function must be used.

4

2 回答 2

3

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
于 2013-09-19T08:09:14.303 回答
0

尝试这个

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
于 2013-09-19T08:07:42.683 回答