编写 HAVING 条件的最有效方法答案

【问题标题】：Most Efficient way to write HAVING condition编写 HAVING 条件的最有效方法
【发布时间】：2015-06-18 18:35:10
【问题描述】：

我目前正在完成sqlzoo的More JOIN operations教程，遇到以下代码作为#12的答案：

SELECT yr,COUNT(title) 
FROM movie 
JOIN casting 
JOIN actor   
ON actorid=actor.id AND movie.id=movieid
WHERE name='John Travolta'
GROUP BY yr

HAVING COUNT(title)=
(SELECT MAX(c) FROM
(SELECT yr, COUNT(title) AS c 
FROM movie 
JOIN casting ON movie.id=movieid
JOIN actor   ON actorid=actor.id
WHERE name='John Travolta'
GROUP BY yr) AS t
)

没有更简洁的方式来表达这段代码吗？

【问题讨论】：

“对于他制作超过 2 部电影的任何年份。” 所以HAVING COUNT(title)>2

标签： mysql sql

【解决方案1】：

是的。如果我正确理解了这个问题，那就简单多了。

SELECT yr,COUNT(title) 
   FROM movie JOIN casting ON movie.id=movieid
        JOIN 
        actor   ON actorid=actor.id
WHERE name='John Travolta'
GROUP BY yr
HAVING COUNT(title) > 2

【讨论】：

【解决方案2】：

看起来原始网站的建议答案与他们提出的问题不符。

他们所问问题的最佳答案：

SELECT yr,COUNT(title) FROM
  movie JOIN casting ON movie.id=movieid
         JOIN actor   ON actorid=actor.id
WHERE name='John Travolta'
GROUP BY yr
HAVING COUNT(title)>2

他们的回答是：

SELECT yr,COUNT(title) FROM
movie JOIN casting ON movie.id=movieid
JOIN actor ON actorid=actor.id
WHERE name='John Travolta'
GROUP BY yr
HAVING COUNT(title)=
(SELECT MAX(c) FROM (SELECT yr,COUNT(title) AS c 
FROM movie 
JOIN casting ON movie.id=movieid JOIN actor ON actorid=actor.id
WHERE name='John Travolta'
GROUP BY yr) AS t

)

但这实际上回答了一个不同的问题，即：“在他制作电影最多的那一年，那是哪一年，他制作了多少部电影？”

【讨论】：