Question

所以我在postgreSQL（以及其他）中提交了提交，用户，故事和视图表，如下所示：

submissions
-----------
user_id
story_id
body

views
-----------
user_id
submissions_id

users
-----------
username

我有一个复杂的SQL查询查找，它返回按用户是否看到它排序的提交（在views表中有user_id和submission_id的条目），然后是它有多少个视图：

    submissions.find_by_sql("select submissions.*,
          count(views.submission_id) as views_count,
          (select count(*) from views
            where views.user_id=#{user_id} and views.submission_id=submissions.id group by id) as counted
          from submissions
          left outer join views on views.submission_id=submissions.id
          where submissions.story_id = #{id}
          group by submissions.id
          order by counted desc, views_count")

我还希望能够获得每次提交的用户名，这应该很简单：只需在users.id = submissions.user_id加入users表，并将users.username添加到选择部分。但是，当我这样做时，我收到错误

column "users.username" must appear in the GROUP BY clause or be used in an aggregate function

这不应该，因为将用户添加到分组依据会删除具有相同用户的多次提交，并且我不需要聚合函数。

由于我显然必须明确说明这一点，我知道还有很多其他问题要问这个问题，而且没有一个问题能真正回答我的问题。

Answer 1

当您拥有group by子句时，在大多数系统中，您都被限制为仅显示group by语句中select子句中的字段。加上使用count(), sum(),等的任何聚合列

尝试围绕原始查询的其他查询：

select * from 
(
select top 100 percent 
          submissions.id,
          count(views.submission_id) as views_count,
          (select count(*) from views
            where views.user_id=#{user_id} and views.submission_id=submissions.id group by id) as counted
          from submissions
          left outer join views on views.submission_id=submissions.id
          where submissions.story_id = #{id}
          group by submissions.id
          order by counted desc, views_count
) as InnerQuery
left join submissions on submissions.ID = InnerQuery.ID

column必须出现在GROUP BY子句中或用于聚合函数

1 个答案: