Question

我表格中的数据如下所示

PAY_END_DT  Sal
10/27/2013  0
11/10/2013  0
11/24/2013  2473.14
12/08/2013  0
01/19/2014  0
02/02/2014  0
02/16/2014  0

我想要的结果应如下所示

10/27/2013 11/10/2013
12/08/2013 02/16/2014

我需要一个SQL来生成这个结果集..请帮助

Answer 1

SELECT
    employee_id,
    MIN(pay_end_dt)   AS island_min_pay_end_dt,
    MAX(pay_end_dt)   AS island_max_pay_end_dt
FROM
(
    SELECT
        pay_end_dt,
        ROW_NUMBER() OVER (PARTITION BY employee_id,
                               ORDER BY pay_end_dt   )     AS full_set_ordinal,
        ROW_NUMBER() OVER (PARTITION BY employee_id, sal
                               ORDER BY pay_end_dt   )     AS zero_set_ordinal
    FROM
        yourTable
)
    AS sorted_set
WHERE
    sal = 0
GROUP BY
    employee_id,
    full_set_ordinal - zero_set_ordinal
;

以您的数据为例：

PAY_END_DT  Sal      FULL_SET_ORIDINAL  ZERO_SET_ORDINAL  "FULL - ZERO"
10/27/2013  0        1                  1                 0
11/10/2013  0        2                  2                 0
11/24/2013  2473.14  3                  1                 2
12/08/2013  0        4                  3                 1
01/19/2014  0        5                  4                 1
02/02/2014  0        6                  5                 1
02/16/2014  0        7                  6                 1

然后，我们只允许包含行WHERE sal = 0然后GROUP BY "FULL - ZERO"来获取我们的两个集合，然后最后应用MIN()和MAX()函数。

正如评论中所述，这被称为＆＃34; Gaps and Islands＆＃34;。

    1 1 1   1   
0 0       0   0 0

1 2 3 4 5 6 7 8 9   - ordinal from the whole set

    1 2 3   4       - ordinal from just the "islands"
1 2       3   4 5   - ordinal from just the "gaps"

    2 2 2   3       - ordinal of the "islands"  (whole_set_id - islands_id)
0 0       3   4 4   - ordinal of the "gaps"     (whole_set_id - gaps_id)

Answer 2

;）

    SELECT "10/27/2013 11/10/2013"
    UNION  
    SELECT "12/08/2013 02/16/2014"

从SQL Server SQL获取最小和最大日期

2 个答案: