找出所有与 Yash Chopra 合作的电影比任何其他导演都多的演员答案

【问题标题】：Find all the actors that made more movies with Yash Chopra than any other director找出所有与 Yash Chopra 合作的电影比任何其他导演都多的演员
【发布时间】：2022-01-13 13:29:07
【问题描述】：

Scehma

    SELECT   p1.pid, 
         p1.NAME, 
         Count(movie.mid) AS movieswithyc 
FROM     person           AS p1 natural 
JOIN     m_cast natural 
JOIN     movie 
JOIN     m_director 
ON       ( 
                  movie.mid = m_director.mid) 
JOIN     person AS p2 
ON       ( 
                  m_director.pid = p2.pid) 
WHERE    p2.NAME LIKE 'Yash Chopra' 
GROUP BY p1.pid 
HAVING   Count(movie.mid) >ALL 
         ( 
                    SELECT     Count(movie.mid) 
                    FROM       person AS p3 natural 
                    JOIN       m_cast 
                    INNER JOIN movie 
                    JOIN       m_director 
                    ON         ( 
                                          movie.mid = m_director.mid) 
                    JOIN       person AS p4 
                    ON         ( 
                                          m_director.pid = p4.pid) 
                    where      p1.pid = p3.pid 
                    AND        p4.NAME NOT LIKE 'Yash Chopra' 
                    GROUP BY   p4.pid) 
ORDER BY movieswithyc DESC;

我没有得到正确的输出。我得到零行。有人可以修改上面的查询并给我正确的输出吗，我尝试了各种查询但没有得到任何东西

【问题讨论】：

这个也不行
可以清楚地解释您要达到的目标。或者可能发布您的数据和预期结果
我正在使用 sqlite ，并且在该查询之后我得到零行。
你能清楚地展示表格的数据和预期的结果吗？

标签： sql

【解决方案1】：

检查一下：

SELECT first.actor, 
       first.count 
FROM   (SELECT Trim(actor) AS Actor, 
               Count(*)    AS COUNT 
        FROM   m_cast mc 
               INNER JOIN (SELECT m.mid 
                           FROM   movie m) AS m 
                       ON m.mid = Trim(mc.mid) 
               INNER JOIN (SELECT md.pid, 
                                  md.mid 
                           FROM   m_director md) AS md 
                       ON md.mid = Trim(mc.mid) 
               INNER JOIN (SELECT p.pid, 
                                  p.NAME AS actor 
                           FROM   person p) AS pactor 
                       ON pactor.pid = Trim(mc.pid) 
               INNER JOIN (SELECT p.pid, 
                                  p.NAME AS director 
                           FROM   person p) AS pdirector 
                       ON pdirector.pid = Trim(md.pid) 
        WHERE  director LIKE '%Yash Chopra%' 
        GROUP  BY Trim(actor)) first 
       LEFT JOIN (SELECT actor, 
                         Max(count) AS COUNT 
                  FROM   (SELECT DISTINCT Trim(actor) AS Actor, 
                                          Count(*)    AS COUNT 
                          FROM   m_cast mc 
                                 INNER JOIN (SELECT m.mid 
                                             FROM   movie m) AS m 
                                         ON m.mid = Trim(mc.mid) 
                                 INNER JOIN (SELECT md.pid, 
                                                    md.mid 
                                             FROM   m_director md) AS md 
                                         ON md.mid = Trim(mc.mid) 
                                 INNER JOIN (SELECT p.pid, 
                                                    p.NAME AS actor 
                                             FROM   person p) AS pactor 
                                         ON pactor.pid = Trim(mc.pid) 
                                 INNER JOIN (SELECT p.pid, 
                                                    p.NAME AS director 
                                             FROM   person p) AS pdirector 
                                         ON pdirector.pid = Trim(md.pid) 
                          WHERE  director NOT LIKE '%Yash Chopra%' 
                          GROUP  BY Trim(actor), 
                                    director) 
                  GROUP  BY actor) second 
              ON first.actor = second.actor 
WHERE  first.count >= second.count 
        OR second.actor IS NULL 
ORDER  BY first.count DESC

【讨论】：

【解决方案2】：

您可以检查以下 SQL。

说明 - 第一个内联视图返回带有“Yash Chopra”的电影数量的人员列表。第二个内联视图返回与其他导演合作的电影数量的人员列表。最后，我筛选出那些使用“Yash Chopra”的电影数量大于“其他导演”的人。

(select pc.name, count(distinct m.mid) count_movie
from movie m
join m_cast mc on m.mid = mc.mid
join m_director md on m.mid = md.mid
join person pc on mc.pid = pc.pid
join person pd on md.pid = pd.pid
where pd.name = 'YASH CHOPRA'
group by pc.name) lst_yc
join
(select pc.name, count(m.mid) count_movie
from movie m 
join m_cast mc on m.mid = mc.mid
join m_director md on m.mid = md.mid
join person pc on mc.pid = pc.pid
join person pd on md.pid = pd.pid
where pd.name != 'YASH CHOPRA'
group by pc.name) lst_wo
on lst_yc.name = lst_wo.name
where lst_yc.count_movie > lst_wo.count_movie

【讨论】：

这个答案在错误的上下文中。在这里，您将 Yash Chopra 与其他导演的总和进行比较。这不是问题上下文。您正在这样做：actor1, director1 ->15 actor1, director2 ->8 actor1, director3 ->9 假设导演 1 是 Yash Chopra 所以，根据问题上下文答案是：actor1, director1 ->15 因为它是 Yash Chopra 的最高值，但根据您的查询，它将比较像 15 > 8+9 这不是真的，所以它会返回一个空集，根据问题的上下文是错误的。

【解决方案3】：

 SELECT * 
FROM   ( 
                SELECT   pc.NAME, 
                         Count(DISTINCT Trim(m.mid)) count_movie 
                FROM     movie m 
                JOIN     m_cast mc 
                ON       Trim(m.mid) = Trim(mc.mid) 
                JOIN     m_director md 
                ON       Trim(m.mid) = Trim(md.mid) 
                JOIN     person pc 
                ON       Trim(mc.pid) = Trim(pc.pid) 
                JOIN     person pd 
                ON       trim(md.pid )= Trim(pd.pid) where pd.NAME = 'Yash Chopra' GROUP BY pc.NAME) lst_yc
                JOIN 
                         ( 
                                  SELECT   pc.NAME, 
                                           count(trim(m.mid)) count_movie 
                                  FROM     movie m 
                                  JOIN     m_cast mc 
                                  ON       trim(m.mid) = trim(mc.mid ) 
                                  JOIN     m_director md 
                                  ON       trim(m.mid) = (md.mid) 
                                  JOIN     person pc 
                                  ON       trim(mc.pid) = trim(pc.pid) 
                                  JOIN     person pd 
                                  ON       trim(md.pid) = trim(pd.pid) 
                                  WHERE    pd.NAME != 'Yash Chopra' 
                                  GROUP BY pc.NAME) lst_wo 
                ON       lst_yc.NAME = lst_wo.NAME 
                WHERE    lst_yc.count_movie > lst_wo.count_movie

这似乎是山塔努先生给出的答案。但是你知道为什么这需要时间吗，我在 1 小时前运行了查询，但还没有产生结果。

【讨论】：

你能把电影，人物和其他表中的记录数等详细信息吗？您使用的是哪个数据库？它是关系数据库还是nosql db？请在您的原始问题中提供尽可能多的详细信息。
我正在使用 sqlite ，并且在查询@ShantanuKher 之后我得到零行

【解决方案4】：

p2.NAME LIKE 'Yash Chopra' 和 p1.PID 这是您的代码行。你应该这样写 TRIM(p2.NAME),TRIM(p1.PID) 因为电影表中的名称和 PID 包含空格和类似的东西。你应该正确处理它，否则它将返回零行，保留它记在心里。

【讨论】：

【解决方案5】：

select t.actor,t.count from ( SELECT actor,count(distinct m.mid) as count
                      FROM   m_cast mc 
                             INNER JOIN (SELECT m.mid 
                                         FROM   movie m) AS m 
                                     ON m.mid = Trim(mc.mid) 
                             INNER JOIN (SELECT md.pid, 
                                                md.mid 
                                         FROM   m_director md) AS md 
                                     ON md.mid = Trim(mc.mid) 
                             INNER JOIN (SELECT p.pid, 
                                                p.NAME AS actor 
                                         FROM   person p) AS pactor 
                                     ON pactor.pid = Trim(mc.pid) 
                             INNER JOIN (SELECT p.pid, 
                                                p.NAME AS director 
                                         FROM   person p) AS pdirector 
                                     ON pdirector.pid = Trim(md.pid) 
                      WHERE  director   LIKE '%Yash Chopra%' 
                      --and actor like '%Uttam Sodi%'
                      group by actor) as t
                      join( SELECT actor,count(distinct m.mid) as count
                      FROM   m_cast mc 
                             INNER JOIN (SELECT m.mid 
                                         FROM   movie m) AS m 
                                     ON m.mid = Trim(mc.mid) 
                             INNER JOIN (SELECT md.pid, 
                                                md.mid 
                                         FROM   m_director md) AS md 
                                     ON md.mid = Trim(mc.mid) 
                             INNER JOIN (SELECT p.pid, 
                                                p.NAME AS actor 
                                         FROM   person p) AS pactor 
                                     ON pactor.pid = Trim(mc.pid) 
                             INNER JOIN (SELECT p.pid, 
                                                p.NAME AS director 
                                         FROM   person p) AS pdirector 
                                     ON pdirector.pid = Trim(md.pid) 
                      WHERE  director not  LIKE '%Yash Chopra%' 
                      group by actor) as w
                where t.actor=w.actor and t.count>=w.count

【讨论】：

【解决方案6】：

大家好，sql 新手，像我一样努力解决这个问题，你可以在下面找到解决方案的一部分（99%），因为我不想打断你的学习过程。但在经历它之前，请尝试最后一次。我感谢在这个论坛上讨论过他们对这个问题的各种想法的人们，因为他们引发了我的各种想法。

在完成解决方案之前，您可以查看video 以了解以下代码中使用的各种新关键字的概述。免责声明 - 在需要时使用修剪选项

select actor,movies from 
    
( select mc.pid as actor,
         md.pid as director,
         p.pid,
         count(*) as movies,
         rank() over (partition by mc.pid order by count(*) desc) as rn,
         p.name
  from m_director as md 
  join
  m_cast as mc on md.mid=mc.mid
  left join
  person as p on md.pid=p.pid and name = 'Yash Chopra'
  group by mc.pid,md.pid
 )
    
where rn =1 and director like "nm0007181" ;

精确解决方案 - 为了获得精确解决方案，您可以将上述表格与 people 表格连接起来，以获得 演员的姓名 yash Chopra 导演的次数比其他任何导演都多。

paila saisravan - 数据挖掘者

【讨论】：

【解决方案7】：

select p.name,h.count 
from(select mc.pid as mcpid,md.pid as mdpid,count(mc.MID) as count   
from m_cast as mc
                 join m_director md 
                     on md.MID=mc.MID               
                 group by mc.pid ,md.pid 
                ) h
            join person p 
                on h.mcpid=p.pid
            where h.count = (select count(*) as count   
                             from m_cast as mc
                             join m_director md 
                                 on md.mid=mc.mid
                             where mc.pid=h.mcpid 
                             group by mc.pid,md.pid 
                             order by count(*) desc
                             limit 1)
            and h.mdpid = (select pid 
                           from person 
                           where name like '%Yash Chopra%'
                          )
            order by h.count desc

【讨论】：