通过比较子集选择记录

时间:2012-07-11 13:14:13

标签: sql sql-server-2005

给定两个表(每个表中的行是不同的):

 1)  x | y     z   2)  x | y     z
    -------   ---     -------   ---
     1 | a     a       1 | a     a
     1 | b     b       1 | b     b
     2 | a             1 | c
     2 | b             2 | a
     2 | c             2 | b
                       2 | c

有没有办法选择第一个表的x列中的值,y列中x列的值子集与z列的值完全匹配在第二个表的1)列中?

如果1,预期结果为c。如果将2添加到第二个表格,则预期结果为2)no record的情况下,预期结果为c,因为第一个表中的子集都不匹配第二个表中的子集。如果将1, 2添加到第二个表格,则预期结果为except

我已尝试使用intersectintersect将第一个表的子集与第二个表进行比较,这样可以正常工作,但{{1}}部分需要太长时间,我可以弄清楚原因(第一张表有大约10.000条记录,第二张大约有10条记录)。

编辑:我已经更新了问题以提供额外的方案。

2 个答案:

答案 0 :(得分:7)

SELECT
  table1.x
FROM
  table1
INNER JOIN
  table2
    ON table1.y = table2.z
GROUP BY
  table1.x
HAVING
      COUNT(*) = (SELECT COUNT(*) FROM table2 AS lookup)
  AND COUNT(*) = (SELECT COUNT(*) FROM table1 AS lookup WHERE x = table1.x)

答案 1 :(得分:1)

其中一个会做

select
    t1.x
from 
    table1 as t1 inner join table2 as t2 on t1.x=t2.x
group by t1.x
having count(distinct t1.x)=count(distinct t2.x)

select
    t1.x
from 
    table1 as t1 inner join table2 as t2 on t1.x=t2.x
group by t1.x
having count(distinct t1.x)=(select count(distinct x) from table2)
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