PostgreSQL 如何使用 with as答案

【问题标题】：PostgreSQL how to use with asPostgreSQL 如何使用 with as
【发布时间】：2020-12-09 05:09:56
【问题描述】：

有人知道为什么这不起作用吗？我得到：错误：“most_recent”或附近的语法错误

with most_recent as (SELECT MAX(public."Master_playlist".updated_at)
FROM public."Master_playlist")
SELECT * from public."Playlist"
JOIN public."Master_playlist_playlist" on public."Playlist".id = public."Master_playlist_playlist".playlist_id
JOIN public."Master_playlist" on public."Master_playlist_playlist".master_playlist_id = public."Master_playlist".id
WHERE public."Master_playlist".updated_at = most_recent;

应该从 Master_playlist 获取最新日期，然后使用它来选择 Master_playlist 以加入内部查询

谢谢！嗯

【问题讨论】：

标签： sql postgresql sql-order-by where-clause greatest-n-per-group

【解决方案1】：

with 子句创建一个派生表，您需要从中选择，使用 join 或子查询。您还需要为该列设置别名，以便以后可以引用它，如下所示：

with most_recent as (
    SELECT MAX(updated_at) max_updated_at
    FROM public."Master_playlist"
)
SELECT * 
from public."Playlist"
JOIN public."Master_playlist_playlist" 
    on public."Playlist".id = public."Master_playlist_playlist".playlist_id
JOIN public."Master_playlist" 
    on public."Master_playlist_playlist".master_playlist_id = public."Master_playlist".id
WHERE public."Master_playlist".updated_at = (SELECT max_updated_at FROM most_recent)

但是在这里，使用行限制查询似乎更简单：

select ...
from (
    select *
    from public."Master_playlist"
    order by updated_at desc
    limit 1
) mp
inner join public."Master_playlist_playlist" mpp
    on mpp.master_playlist_id = mp.id
inner join public."Playlist" p
    on p.id = mpp.playlist_id

【讨论】：