Question

最近，我看到一位用户，他是唯一获得一些徽章的用户。

奇怪的是，我前往data.se寻找类似的例子（data.se - Single awarded badges）

select count(*) as awarded,
       name, class
from badges
group by name, class
having count(*) = 1

更进一步，我还希望看到拥有这些徽章的用户（data.se - Count awarded badges and users）

select count(*) over (partition by b.name, b.class) as awarded,
       b.name, b.class, u.displayname
from badges b
join users u on u.id = b.userid

这给了我所有条目，而不仅仅是单个条目，例如： count(*) = 1。要获得这些，我必须使用CTE或子查询，例如（data.se - Single awarded badges and users）

with q as (select count(*) over (partition by b.name, b.class) as awarded,
                  b.name, b.class, u.displayname
           from badges b
           join users u on u.id = b.userid)
select name, class, displayname
from q
where awarded = 1

最后两个查询的缺点是性能。（非缓存）查询大约需要简单group by/having查询的20倍。使用子查询没有任何改进。

我知道，您无法使用window function in a where or having clause，也无法在第二个查询中说where awarded = 1。

是否有更快更简单的查询，甚至可能没有CTE或子查询？我给一个（用CTE）作为我自己的答案，但是希望看到替代解决方案。

Answer 1

一种解决方案是将group by/having查询作为CTE并加入徽章和用户表（data.se - Single awarded badges and users）

with awarded as (select count(*) as num,
                        name, class
                 from badges
                 group by name, class
                 having count(*) = 1)
select b.name, b.class, u.displayname
from badges b
join awarded a on a.name = b.name and a.class = b.class
join users u on u.id = b.userid

根据窗口函数过滤行

1 个答案: