Question

从下面具有架构的数据库

movieActor (actorID, movieID)
rental (rentalID, inventoryID, customerID)
inventory (inventoryID, movieID)

我正在尝试列出一对从同一演员租借电影的客户。结果集应由三列组成

customerID1,customerID2,nOfCommonActors

例如

23 44 5 
11 44 3

其中第一行表示ID为23和44的客户分别租借了各种电影，但在这两组电影中都扮演过的演员中有5个客户出租了电影23和44

我想出了这个查询，但是它要花很多时间运行，并且超时而没有返回任何结果。想知道如何提高效率（我正在使用MYSQL）：

SELECT r1.customerID AS customerID1,
    r2.customerID AS customerID2,
    COUNT(DISTINCT fa.actorID) as nOfCommonActors
FROM movieActor AS fa
    JOIN (SELECT r.customerID, i.movieID, fa.actorID
    FROM rental AS r
        JOIN inventory i
        ON i.inventoryID=r.inventoryID
        JOIN movieActor AS fa
        ON fa.actorID=i.movieID
) AS r1
    JOIN (SELECT r.customerID, i.movieID, fa.actorID
    FROM rental AS r
        JOIN inventory i
        ON i.inventoryID=r.inventoryID
        JOIN movieActor AS fa
        ON fa.actorID=i.movieID
) AS r2
    ON r2.actorID=r1.actorID
        AND r1.customerID < r2.customerID 
GROUP BY r1.customerID, r2.customerID
ORDER BY nOfCommonActors DESC;

Answer 1

我能想到的一件事是子查询中的select distinct：

SELECT ca.customerID AS customerID1,
       ca2.customerID AS customerID2,
       COUNT(*) as nOfCommonActors
FROM (SELECT DISTINCT r.customerID, fa.actorID
      FROM rental r JOIN
           inventory i
           ON i.inventoryID = r.inventoryID JOIN
           movieActor fa
           ON fa.actorID = i.movieID
     ) ca JOIN
     (SELECT DISTINCT r.customerID, fa.actorID
      FROM rental r JOIN
           inventory i
           ON i.inventoryID = r.inventoryID JOIN
           movieActor fa
           ON fa.actorID = i.movieID
    ) ca2
    ON ca.actorID = ca2.actorID AND
        ca.customerID < ca2.customerID
GROUP BY ca.customerID, ca2.customerID
ORDER BY nOfCommonActors DESC;

您的版本正在大大增加子查询中的行数。这会使JOIN变得更加昂贵-而且所有多余的工作都变得一无所有，因为您仍然想COUNT(DISTINCT)。

Answer 2

将查询拆分为sectoin，以便统计信息绘制最佳路径

SELECT DISTINCT r.customerID，fa.actorID 进入＃t1 从出租r JOIN 库存我 ON i.inventoryID = r.inventoryID加入 movieActor fa 开启fa.actorID = i.movieID

选择＃t1.customerID AS customerID1，＃t2.customerID AS客户ID2， COUNT（*）作为nOfCommonActors 从（选择＃t1.customerID，＃t2.customerID 从＃t1 加入＃t2 ON＃t1.actorID =＃t2.actorID和＃t1.customerID <＃t2.customerID）

按＃t1.customerID，＃t2.customerID分组通过nOfCommonActors DESC进行订购；

如何优化此查询以防止超时

2 个答案: