Question

在TABLE_A中，每个ID都有一个最小SDT和最大EDT。

示例：对于ID = 1000，这将是1/09/2013和3/10/2013。

示例：对于ID = 1001，这将是1/07/2013和3/08/2013。

预期结果将是来自TABLE_B的所有记录，其中来自TABLE_B的SDT落在来自TABLE_A的那些最小SDT和最大EDT值之间。

TABLE_A

ID      SDT          EDT
------------------------------
1000    1/10/2013    3/10/2013  
1000    1/09/2013    3/09/2013   
1001    1/08/2013    3/08/2013
1001    1/07/2013    3/07/2013
1002    1/06/2013    3/06/2013

表-B

ID      SDT
-----------------
1000    2/10/2013   ===> because it falls between 1/09/2013 and 3/10/2013
1000    4/09/2013   ===> because it falls between 1/09/2013 and 3/10/2013
1001    2/08/2013   ===> because it falls between 1/07/2013  and 3/08/2013
1001    4/05/2013   ==> not in result because it's not between 1/07/2013 and 3/08/2013
1002    4/06/2013   ==> also out of range, so not in results

Answer 1

您可以将group by用于最小值/最大值，将相关子查询用于table_b

Select
    *
From
    Table_b b
Where
    Exists (
        Select
            'x'
        From
            table_a a
        Where
            a.id = b.id
        group by
            a.id
        having
            min(a.sdt) <= b.sdt and
            max(a.edt) >= b.sdt
  )

<强> Example Fiddle

Answer 2

值得注意的是，在子查询中使用GROUP BY会产生不必要的性能成本。

SELECT b.*
FROM table_b b
WHERE EXISTS (
    SELECT 1
    FROM table_a a
    WHERE b.sdt BETWEEN LEAST(a.sdt, e.edt) AND GREATEST(a.edt, a.sdt)
      AND a.id = b.id
);

编辑/校正：
只有在匹配table_a中的各个行而不是MIN / MAX分组时，此解决方案才适用。我以前忽略了table_a的MIN / MAX分组要求。

以下使用两个没有GROUP BY的EXISTS查询并返回正确的结果：

SELECT b.*
FROM table_b b
WHERE (EXISTS (
    SELECT 1
    FROM table_a a
    WHERE b.sdt >= a.sdt 
    AND a.id = b.id
) AND EXISTS (
    SELECT 1
    FROM table_a a
    WHERE b.sdt <= a.edt 
    AND a.id = b.id
));

ORACLE SQL查询记录日期在其他表的最小和最大日期之间

2 个答案: