SQL Select 每个用户的最新 (public) post

Question

数据示例可能如下所示：

用户名	推文	is_public	created_at
卡特莉亚	tweet1	1	2021-06-14
你	tweet1	1	2021-06-15
尼克	头尾	1	2021-06-15
你	lasttweet	1	2021-06-23
尼克	空	0	2021-06-23

到select所有public推文:

 WHERE is_public = 1

至 select 每个用户的最新推文 我受到 these 的启发，所以我得到：

SELECT posts.* FROM post
  LEFT JOIN posts as postdouble
    ON posts.created_at < postdouble.created_at AND posts.username = postdouble.username
WHERE postdouble.username IS NULL

每个用户select最新和public推文（混淆这两个陈述）我' m 被一些 SubSubQueries

丢失了

SELECT posts.username, posts.tweet, posts.is_public, posts.created_at, postdouble.username as userd, postdouble.tweet as tweetd, postdouble.is_public as ispublicd, postdouble.created_at as createdatd FROM posts
  LEFT JOIN posts as postdouble
    ON posts.created_at < (
         SELECT MAX(posts.created_at) FROM posts WHERE is_public = 1
      )
    AND posts.username = postdouble.username
-- WHERE postdouble.username IS NULL

urs selected 正确（双打中的空实体）但这是唯一正确的事情。请帮我一些debugging.thx^^lg

Answer 1

你把这个复杂化了。您可以使用相关子查询，但不需要额外的 JOIN:

SELECT p.*
FROM posts p
WHERE p.created_at = (SELECT MAX(p2.created_at)
                      FROM posts p2
                      WHERE p2.is_public = 1 AND
                            p2.username = p.username
                     ) AND
      p.is_public = 1;

关联子句确保它是针对同一用户的。

为了获得最佳性能，您需要在 posts(username, public, created_at) 上建立索引。

Answer 2

另一种简单的方法是使用 window function 来选择所需的行

select * from (
    select *, Row_Number() over(partition by username order by created_at desc) rn
    from posts
    where is_public=1
)p
where rn=1

在这种情况下，您也可以使用 top 1 with ties

获得相同的结果

select top (1)  with ties *, Row_Number() over(partition by username order by created_at desc) rn
from posts
where is_public=1
order by rn

SQL Select 每个用户的最新 (public) post

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