3

我在 MySQL 中有以下关系：

Suppliers(**sid**; sname; city),
Parts(**pid**; pname; color),
Catalog(**sid**; **pid**; cost),

键在 *.

我想找到：

  1. 供应商对 (s1; s2) 使得 s1 和 s2 提供完全相同的零件。
  2. 提供至少两个其他人不提供的零件的供应商。
4

3 回答 3

0

非常感谢您的回复。我自己写了以下查询，但我不确定它是否正确。请告诉我。谢谢。

1)

选择 T1.sid，T2.sid

从 T1、T2

哪里（选择 T1.sid，C2.pid

   FROM( SELECT c1.pid

       FROM Catalog C1

       GROUP BY c1.sid) AS T1, Catalog C2

  WHERE T1.sid=c2.sid)

=

(SELECT   T2.sid, C4.pid

   FROM( SELECT c3.pid

       FROM Catalog C3

       GROUP BY c3.sid) AS T2, Catalog C4

  WHERE T2.sid=c4.sid)

2)

选择 T1.S

从 T1

WHERE NOT EXISTS(( SELECT c1.sid AS S, c1.pid AS p1, C2.pid AS p2

           FROM catalog AS C1, Catalog as C2

            WHERE c1.sid=c2.sid AND

            C1.pid<>c2.pid) AS T1

        EXCEPT

        ( SELECT c3.sid AS S1, c3.pid AS p3, C4.pid AS p4

           FROM catalog AS C3, Catalog as C4

           WHERE c3.sid=c4.sid AND

            c3.pid<>c4.pid) AS T2)

不存在（（选择 c3.sid 作为 S1，c3.pid 作为 p3，C4.pid 作为 p4

           FROM catalog AS C3, Catalog as C4

           WHERE c3.sid=c4.sid AND

            c3.pid<>c4.pid) AS T2)

        EXCEPT

        (SELECT c1.sid AS S, c1.pid AS p1, C2.pid AS p2

           FROM catalog AS C1, Catalog as C2

            WHERE c1.sid=c2.sid AND

            C1.pid<>c2.pid) AS T1
于 2012-10-09T17:37:55.770 回答
0

1. 供应商对 (s1; s2) 使得 s1 和 s2 供应完全相同的零件：

select c1.sid,c2.sid
from catalog c1 join catalog c2 on c1.pid = c2.pid and c1.sid < c2.sid 
          and not exists 
          (
             select 1 
             from catalog c3 join catalog c4 
             on c3.sid = c1.id and c4.id = c2.sid 
                               and c3.pid <> c4.pid
          )

2.供应商至少提供两个其他人不提供的零件。

select c1.sid
from catalog c1 join catalog c2 on c1.sid = c2.sid and c1.pid < c2.pid 
          and not exists 
          (
             select 1 
             from catalog  
             where sid <> c1.sid and (pid = c1.pid or pid = c2.pid)
          )
于 2012-10-09T14:43:24.767 回答
0 
SELECT supplier_1, supplier_2
FROM (  SELECT c.sid AS supplier_1, c2.sid AS supplier 2, COUNT(*) AS total, COUNT(c2.sid) AS connected
        FROM Catalog AS c
        LEFT JOIN Catalog AS c2
        ON c.sid != c2.sid AND c.pid = c2.pid
        GROUP BY c.sid, c2.sid) AS h
GROUP BY LEAST(supplier_1, supplier_2), GREATEST(supplier_1, supplier_2)

我想像这样的事情会成功。基本上只是LEFT JOIN根据不等于同一个供应商和相等的部分id做一个，然后看看连接的数量是否等于没有成功连接的行数。

于 2012-10-09T14:44:43.440 回答