如何獲取唯一值來比較同一張表的兩列？

Question

目前，我正在使用 SQLite 處理 Django 項目，在該項目中我創建了一個消息聊天框，並且我想獲取登錄用戶的最新消息，無論它是發送還是接收。

id   receiver_id      sender_id    message_content   created_at
1    1                 2            some text         2020-08-11 13:29:47.342944
3    3                 2            some text         2020-08-11 13:44:55.499638
4    2                 1            some text         2020-08-11 14:20:55.499638
5    1                 2            some text         2020-08-12 05:06:05.497500
6    2                 5            some text         2020-08-12 10:39:31.234082
7    4                 1            some text         2020-08-14 13:25:19.357876

使用下面 SQL 查詢后。

SELECT max(created_at), * 
FROM hireo_messages 
WHERE receiver_id=2 or sender_id=2 
GROUP BY receiver_id, sender_id 
ORDER BY created_at DESC

我得到了以下結果。

id      receiver_id   sender_id     message_content   created_at
6       2             5             some text         2020-08-12 10:39:31.234082
5       1             2             some text         2020-08-12 05:06:05.497500
4       2             1             some text         2020-08-11 14:20:55.499638
3       3             2             some text         2020-08-11 13:44:55.499638

如您所見，id 5、4 都在互相聊天。 所以我想獲取除 id 4 之外的所有記錄，因為 2 個用戶之間的最新聊天在 id 5 中。它與 Facebook 信使儀表板中使用的概念相同。 請指導我是通過查詢解決還是使用任何其他方式。 在此先感謝您的幫助！

Answer 1

我想你想要：

SELECT LEAST(receiver_id, sender_id), GREATEST(receiver_id, sender_id), MAX(created_at)
FROM hireo_messages
WHERE 2 IN (receiver_id, sender_id)
GROUP BY LEAST(receiver_id, sender_id), GREATEST(receiver_id, sender_id)
ORDER BY MAX(created_at) DESC;

如果您想要完整的行，請使用 window 函數：

SELECT m.*
FROM (SELECT m.*,
             ROW_NUMBER() OVER (PARTITION BY LEAST(receiver_id, sender_id), GREATEST(receiver_id, sender_id) ORDER BY created_at DESC) as seqnum
      FROM hireo_messages m
      WHERE 2 IN (receiver_id, sender_id)
     ) m
WHERE seqnum = 1
ORDER BY MAX(created_at) DESC;

Answer 2

您可以使用 window function MAX() ：

select id, receiver_id, sender_id, message_content, created_at 
from (
  select *,
    max(created_at) over (partition by min(receiver_id, sender_id), max(receiver_id, sender_id)) last_date
  from hireo_messages
  where 2 in (receiver_id, sender_id) 
)
where created_at = last_date
order by id desc

如果您的 SQLite 版本不支持 window 功能，您可以使用NOT EXISTS ：

select h.* from hireo_messages h
where 2 in (receiver_id, sender_id)
and not exists (
  select 1 from hireo_messages
  where min(receiver_id, sender_id) = min(h.receiver_id, h.sender_id)
    and max(receiver_id, sender_id) = max(h.receiver_id, h.sender_id)
  and created_at > h.created_at
)  
order by h.id desc 

請參閱演示。
結果：

| id  | receiver_id | sender_id | message_content | created_at                 |
| --- | ----------- | --------- | --------------- | -------------------------- |
| 6   | 2           | 5         | some text       | 2020-08-12 10:39:31.234082 |
| 5   | 1           | 2         | some text       | 2020-08-12 05:06:05.497500 |
| 3   | 3           | 2         | some text       | 2020-08-11 13:44:55.499638 |