查询以获取用户收件箱的最后对话

Question

我需要特定的SQL查询来选择用户收件箱的最后10个对话。 收件箱仅向每个用户显示对话（线索） - 它会选择对话中的最后一条消息并将其显示在收件箱中。

已编辑。

预期结果：从10 latest conversations的每一个中提取latest message。 Facebook以同样的方式显示最新对话

还有一个问题。如何使pagination显示下一页中之前最近一次对话的后10条最新消息？

数据库悄悄话的样子：

| id | user_id | recipient_id | text 
| 1 | 2  | 3   | Hi John! 
| 2 | 3  | 2   | Hi Tom! 
| 3 | 2  | 3   | How are you? 
| 4 | 3  | 2   | Thanks, good! You? 

Answer 1

按我的理解，你需要得到每个用户为基础的谈话最新消息（最后10个最新的谈话）

更新：我已经修改了查询来获取该latest_conversation_message_id为每个用户通话

下面的查询获取详细信息user_id = 2，您可以修改，users.id = 2得到它的任何其他用户

SQLFiddle，希望这能解决你的目的

SELECT 
    user_id, 
    users.name, 
    users2.name as sent_from_or_sent_to, 
    subquery.text as latest_message_of_conversation 
FROM 
    users 
    JOIN 
    (
    SELECT 
     text, 
     row_number() OVER (PARTITION BY user_id + recipient_id ORDER BY id DESC) AS row_num, 
     user_id, 
     recipient_id, 
     id 
    FROM 
     private_messages 
    GROUP BY 
     id, 
     recipient_id, 
     user_id, 
     text 
    ) AS subquery ON ((subquery.user_id = users.id OR subquery.recipient_id = users.id) AND row_num = 1) 
    JOIN users as users2 ON (users2.id = CASE WHEN users.id = subquery.user_id THEN subquery.recipient_id ELSE subquery.user_id END) 
WHERE 
    users.id = 2 
ORDER BY 
    subquery.id DESC 
LIMIT 10 

信息：查询得到与任何其他用户每次谈话的最新消息，如果user_id 2，将消息发送到user_id 3，那也显示，作为它表示对话的开始。显示与任何其他用户的每次对话的最新消息

Answer 2

您的WHERE子句

ORDER BY "ColumnName" [ASC, DESC] 

UNION Description at W3Schools它结合了这2个语句的结果后粘贴。

SELECT "ColumnName" FROM "TableName" 
UNION 
SELECT "ColumnName" FROM "TableName" 

Answer 3

要解决pg中的groupwise-max，可以使用DISTINCT ON。就像这样：

SELECT 
DISTINCT ON(pm.user_id) 
    pm.user_id, 
    pm.text 
FROM 
    private_messages AS pm 
WHERE pm.recipient_id= <my user id> 
ORDER BY pm.user_id, pm.id DESC; 

http://sqlfiddle.com/#!12/4021d/19

要获取最新的X但是我们将有子查询中使用它：

SELECT 
q.user_id, 
q.id, 
q.text 
FROM 
(
SELECT 
DISTINCT ON(pm.user_id) 
    pm.user_id, 
pm.id, 
    pm.text 
FROM 
    private_messages AS pm 
WHERE pm.recipient_id=2 
ORDER BY pm.user_id, pm.id DESC 
) AS q 
ORDER BY q.id DESC 
LIMIT 10; 

http://sqlfiddle.com/#!12/4021d/28

要获取发出和收到的线程：

SELECT 
q.user_id, 
q.recipient_id, 
q.id, 
    q.text 
FROM 
(
SELECT 
DISTINCT ON(pm.user_id,pm.recipient_id) 
    pm.user_id, 
    pm.recipient_id, 
pm.id, 
    pm.text 
FROM 
    private_messages AS pm 
WHERE pm.recipient_id=2 OR pm.user_id=2 
ORDER BY pm.user_id,pm.recipient_id, pm.id DESC 
) AS q 
ORDER BY q.id DESC 
LIMIT 10; 

http://sqlfiddle.com/#!12/4021d/42

Answer 4

对于大型数据集我想你可能想尝试运行两个语句，然后巩固成果，作为（USER_ID和id）或（recipient_id和id）索引扫描应该是非常有效的获得每种类型的最近10次会话。

with sent_messages as (
    SELECT * 
    FROM  private_messages 
    WHERE user_id = my_user_id 
    ORDER BY id desc 
    LIMIT 10), 
received_messages as ( SELECT * 
    FROM  private_messages 
    WHERE recipient_id = my_user_id 
    ORDER BY id desc 
    LIMIT 10), 
all_messages as (
    select * 
    from sent_messages 
    union all 
    select * 
    from received_messages) 
select * 
from  all_messages 
order by id desc 
limit 10 

编辑：其实另一个查询值得尝试的可能是：

select * 
from private_messages 
where id in (
     select id 
     from (
      SELECT id 
      FROM  private_messages 
      WHERE user_id = my_user_id 
      ORDER BY id desc 
      LIMIT 10 
      union all 
      SELECT id 
      FROM  private_messages 
      WHERE recipient_id = my_user_id 
      ORDER BY id desc 
      LIMIT 10) all_ids 
      order by id desc 
      limit 10) last_ten_ids 
order by id desc 

这可能是9.2+，其中单独的指标可以被用来获取ID的更好，或者在情况下，最最近检索的数量非常大。尽管如此，还是有点不清楚。如果有疑问，我会去为前一个版本。