Question

我正在解决有关SPJ数据库的一些问题。

S(SNUM, SNAME) [supplier]
P(PNUM, PNAME) [part]
J(JNUM, JNAME) [project]
SPJ(SNUM, PNUM, JNUM, Q)

查询是：获得至少有100个部分的项目对。
到目前为止，我已经为它编写了一个代码：

SELECT X.JNUM, Y.JNUM FROM SPJ AS X, SPJ AS Y
    WHERE 100 <= (SELECT COUNT(PNUM) FROM (SELECT PNUM FROM SPJ WHERE SPJ.JN=X.JN 
                                            INTERSECT
                                           SELECT PNUM FROM SPJ WHERE SPJ.JN=Y.JN));

我认为我的解决方案是正确的。有没有其他方法可以解决这个问题？

Answer 1

只需获得所有相同的部分组合，然后查看每个项目对获得的匹配数：

select spj1.jnum as jnum1, spj2.jnum as jnum2 
from spj spj1
join spj spj2 on spj1.pnum = spj2.pnum and spj1.jnum < spj2.jnum
group by spj1.jnum, spj2.jnum
having count(distinct spj1.pnum) >= 100;

Answer 2

一些想法：

派生表需要别名
由于您正在使用笛卡尔积，因此您希望将联接排除在同一SPJ之外，并且可能您不希望对角线相互镜像，即您不需要已为(SPJ1, SPJ2)和(SPJ2, SPJ1)返回了两对：

SELECT X.JNUM, Y.JNUM FROM SPJ AS X, SPJ AS Y
WHERE 100 <= (SELECT COUNT(PNUM) 
                 FROM (SELECT PNUM FROM SPJ WHERE SPJ.JN=X.JN 
                         INTERSECT
                       SELECT PNUM FROM SPJ WHERE SPJ.JN=Y.JN) as AnAlias)
     AND X.PK > Y.PK; -- Avoid duplicates and self false positives

查询以获得相交

2 个答案: