检索单列唯一值的代表记录

Question

对于 Postgresql 8.x，我有一个包含 (id, user_id, question_id, choice) 的 answers 表，其中 choice 是一个字符串值。我需要一个查询，它将为所有唯一的choice 值返回一组记录（返回所有列）。我正在寻找的是每个独特选择的单一代表性记录。我还希望有一个聚合的votes 列，它是与每条记录附带的每个唯一选择匹配的记录数的count()。我想强制 choice 小写以便进行比较 （HeLLo 和 Hello 应该被视为相等）。我不能GROUP BY lower(choice) 因为我想要结果集中的所有列。按所有列分组会导致返回所有记录，包括所有重复项。

1。离我最近的

select lower(choice), count(choice) as votes from answers where question_id = 21 group by lower(choice) order by votes desc;

问题在于它不会返回所有列。

                     lower                     | votes 
-----------------------------------------------+-------
 dancing in the moonlight                      |     8
 pumped up kicks                               |     7
 party rock anthem                             |     6
 sexy and i know it                            |     5
 moves like jagger                             |     4

2。尝试所有列

select *, count(choice) as votes from answers where question_id = 21 group by lower(choice) order by votes desc;

因为我没有在GROUP BY 中指定SELECT 中的每一列，所以这会引发一个错误，告诉我这样做。

3。指定GROUP BY 中的所有列

select *, count(choice) as votes from answers where question_id = 21 group by lower(choice), id, user_id, question_id, choice order by votes desc;

这只是将所有记录的votes 列转储为1 的表。

如何从 1. 中获取 vote 计数和唯一代表记录，但返回表中的 所有 列？

Answer 1

将分组结果与主表连接起来，然后为每个（问题、答案）组合仅显示一行。

类似这样：

WITH top5 AS (
  select question_id, lower(choice) as choice, count(*) as votes 
  from answers 
  where question_id = 21 
  group by question_id , lower(choice) 
  order by count(*) desc
  limit 5
)
SELECT DISTINCT ON(question_id,choice) *
FROM top5 
JOIN answers USING(question_id,lower(choice))
ORDER BY question_id, lower(choice), answers.id;

Answer 2

这就是我最终得到的结果：

SELECT answers.*, cc.votes as votes FROM answers join (
    select max(id) as id, count(id) as votes 
    from answers
    group by trim(lower(choice))
) cc 
on answers.id = cc.id ORDER BY votes desc, lower(response) asc