我正在尝试编写一个查询(PostgreSQL)以获得“2012年奖项数量最多的电影。”
我有以下表格:
CREATE TABLE Award(
ID_AWARD bigserial CONSTRAINT Award_pk PRIMARY KEY,
award_name VARCHAR(90),
category VARCHAR(90),
award_year integer,
CONSTRAINT award_unique UNIQUE (award_name, category, award_year));
CREATE TABLE AwardWinner(
ID_AWARD integer,
ID_ACTOR integer,
ID_MOVIE integer,
CONSTRAINT AwardWinner_pk PRIMARY KEY (ID_AWARD));
我写了下面的查询,它给出了正确的结果,但我认为有很多代码重复。
select * from
(select id_movie, count(id_movie) as awards
from Award natural join awardwinner
where award_year = 2012 group by id_movie) as SUB
where awards = (select max(count) from
(select id_movie, count(id_movie)
from Award natural join awardwinner
where award_year = 2012 group by id_movie) as SUB2);
所以SUB
和SUB2
是完全相同的子查询。有更好的方法吗?
答案 0 :(得分:7)
您可以使用common table expression来避免代码重复:
with cte_s as (
select id_movie, count(id_movie) as awards
from Award natural join awardwinner
where award_year = 2012
group by id_movie
)
select
sub.id_movie, sub.awards
from cte_s as sub
where sub.awards = (select max(sub2.awards) from cte_s as sub2)
或者您可以使用window function执行此类操作(未经测试,但我认为PostgreSQL允许这样做):
with cte_s as (
select
id_movie,
count(id_movie) as awards,
max(count(id_movie)) over() as max_awards
from Award natural join awardwinner
where award_year = 2012
group by id_movie
)
select id_movie
from cte_s
where max_awards = awards
另一种方法是使用rank()函数(未经测试,可能需要使用两个cte而不是一个):
with cte_s as (
select
id_movie,
count(id_movie) as awards,
rank() over(order by count(id_movie) desc) as rnk
from Award natural join awardwinner
where award_year = 2012
group by id_movie
)
select id_movie
from cte_s
where rnk = 1
更新当我创建此答案时,我的主要目标是展示如何使用cte来避免代码重复。在一般情况下,如果可能的话,最好避免在查询中多次使用cte - 第一次查询使用2次表扫描(或索引搜索)而第二次和第三次只使用一次,所以我应该指定最好使用这些查询。无论如何,@ Erwin在他的回答中做了这个测试。只是为了增加他的重要观点:
natural join
,因为这容易出错。实际上,我的主要RDBMS是SQL Server,它不支持它,所以我更习惯于显式outer/inner join
。awards
和只是过滤来自awardwinner
的行,我不想使用join
,而是使用exists
或in
代替,对我来说似乎更合乎逻辑。with cte_s as (
select
aw.id_movie,
count(*) as awards,
rank() over(order by count(*) desc) as rnk
from awardwinner as aw
where
exists (
select *
from award as a
where a.id_award = aw.id_award and a.award_year = 2012
)
group by aw.id_movie
)
select id_movie
from cte_s
where rnk = 1
答案 1 :(得分:2)
SELECT id_movie, awards
FROM (
SELECT aw.id_movie, count(*) AS awards
,rank() OVER (ORDER BY count(aw.id_movie) DESC) AS rnk
FROM award a
JOIN awardwinner aw USING (id_award)
WHERE a.award_year = 2012
GROUP BY aw.id_movie
) sub
WHERE rnk = 1;
这应该比目前为止的建议更简单,更快捷。使用EXPLAIN ANALYZE
进行测试。
有些情况下,CTE有助于避免代码重复。但不是在这个时间:子查询可以很好地完成工作并且通常更快。
您可以在同一查询级别上运行聚合函数的窗口函数。这就是为什么这样做的原因:
rank() OVER (ORDER BY count(aw.id_movie) DESC) AS rnk
我建议在JOIN条件中使用显式列名而不是NATURAL JOIN
,如果稍后更改/添加列到基础表,则容易发生破坏。
USING
的JOIN条件几乎一样短,但不会轻易破坏。
由于id_movie
不能为NULL(由JOIN条件排除,也是pk的一部分),因此使用count(*)
会更短,速度更快。同样的结果。
更短,更快,但是,如果您只需要一个获胜者:
SELECT aw.id_movie, count(*) AS awards
FROM award a
JOIN awardwinner aw USING (id_award)
WHERE a.award_year = 2012
GROUP BY 1
ORDER BY 2 DESC, 1 -- as tie breaker
LIMIT 1
在此处使用位置参考(1
,2
)作为速记
我将id_movie
添加到ORDER BY
作为决胜局,以防多部电影有资格获胜。
答案 2 :(得分:0)
你不需要这样的东西吗?
SELECT ID_MOVIE, COUNT(*)
FROM AwardWinner
JOIN Award ON Award.ID_AWARD = AwardWinner.ID_AWARD
WHERE award_year = 2012
GROUP BY ID_MOVIE
ORDER BY COUNT(*) DESC
或者可能(取决于你要找的东西):
SELECT ID_MOVIE, COUNT(DISTINCT AwardWinner.ID_AWARD)
FROM AwardWinner
JOIN Award ON Award.ID_AWARD = AwardWinner.ID_AWARD
WHERE award_year = 2012
GROUP BY ID_MOVIE
ORDER BY COUNT(*) DESC