使用按功能分组无法解决问题

Question

任务：按字母顺序获取至少拥有30个主演角色的演员列表。

我的代码：

select name, count(ord=1)
from casting
join actor on actorid=actor.id
where ord=1 and count(ord=1) and exists ( select 1 from casting
 where count(movieid)>=30)
group by actorid,name
order by name

这给我一个错误-无效使用按功能分组。

Answer 1

加入表，按参与者分组，然后将条件放入Haveing子句中。

$ rm $(find . -path './src/**' -name __tests__ | sed -E 's/([^ ]+__tests__)/\1\/*.js \1\/*.js.map/g' | xargs echo rm
rm: cannot remove './src/game/__tests__/*.js': No such file or directory
rm: cannot remove './src/game/__tests__/*.js.map': No such file or directory
rm: cannot remove './src/helpers/__tests__/*.js': No such file or directory
rm: cannot remove './src/helpers/__tests__/*.js.map': No such file or directory

表达式select a.name, sum(case c.ord when 1 then 1 else 0 end) starringroles from actor a inner join casting c on c.actorid = a.id group by a.id, a.name having sum(case c.ord when 1 then 1 else 0 end) >= 30 order by a.name 将计算出主演角色的数量（带有sum(case c.ord when 1 then 1 else 0 end)）。

Answer 2

您不能在where上使用聚合，而需要having

select name, count(*)
from casting
join actor on actorid=actor.id
where ord=1  
and exists ( select 1 from casting
 having count(movieid)>=30)   
 group by actorid,name
 having count(movieid)>=30
 order by name

Answer 3

select MAX(name) AS name, count(*) AS roles
from casting
join actor on actorid=actor.id
group by actorid
HAVING COUNT(*)>=30
order by name;

Answer 4

这是SQLZoo中问题＃13中的一个问题。下图说明了表格：

有关数据库的一些描述：

该数据库具有多对多关系的两个实体（电影和演员）。每个实体都有自己的表。第三个表Cast用来链接它们。这种关系是多对多的，因为每部电影都有许多演员，而且每个演员都出现在许多电影中。

这是我发现有效的答案：

select A.name
from actor A
inner join casting C
on C.actorid = A.id
where  C.ord =1                    /*only the lead roles*/
group by A.id                      /*grouped by Actor ID*/
having count(C.movieid) >=15      /*at least 15 starring roles*/
order by A.name                  /* in alphabetical order*/