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我想获取电子邮件域列表和每个域中的顶级用户。我的方法是对按域分组的每封电子邮件的问题求和，然后使用窗口函数获取顶级用户。但是，这不起作用：

SELECT 
  domain,
  sum(questions_per_email) as questions_per_domain,
  first_value(email) OVER (PARTITION BY domain ORDER BY questions_per_email DESC) as top_user
FROM (
    SELECT email,
           lower(substring(u.email from position('@' in u.email)+1)) as domain,
           count(*) as questions_per_email
      FROM questions q
      JOIN identifiers i ON (q.owner_id = i.id)
      JOIN users u ON (u.identifier_id = i.id)
    GROUP BY email
  ) as per_user
GROUP BY domain, top_user

Postgres 给出以下信息：

ERROR:  column "per_user.questions_per_email" must appear in the GROUP BY clause or be used in an aggregate function
LINE 5: ...t_value(email) OVER (PARTITION BY domain ORDER BY questions_...
                                                             ^

我真的不明白为什么会这样。我很确定应该能够对聚合结果使用窗口函数。请指教！

谢谢，克里斯托弗

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1 回答 1

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您可以像这样更改您的查询：

with cte1 as (
    SELECT email,
           lower(substring(u.email from position('@' in u.email)+1)) as domain
      FROM questions q
      JOIN identifiers i ON (q.owner_id = i.id)
      JOIN users u ON (u.identifier_id = i.id)
), cte2 as (
    select
        domain, email,
        count(*) as questions_per_email,
        first_value(email) over (partition by domain order by count(*) desc) as top_user
    from cte1
    group by email, domain
)
select domain, top_user, sum(questions_per_email) as questions_per_domain
from cte2
group by domain, top_user

sql fiddle demo

于 2013-09-09T06:51:25.747 回答