查詢以獲取用戶收件箱的最後對話

Question

我需要特定的SQL查詢來選擇用戶收件箱的最後10個對話。 收件箱僅向每個用戶顯示對話（線索） - 它會選擇對話中的最後一條消息並將其顯示在收件箱中。

已編輯。

預期結果：從10 latest conversations的每一箇中提取latest message。 Facebook以同樣的方式顯示最新對話

還有一個問題。如何使pagination顯示下一頁中之前最近一次對話的後10條最新消息？

數據庫悄悄話的樣子：

| 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，你需要得到每個用戶爲基礎的談話最新消息（最後10個最新的談話）

更新：我已經修改了查詢來獲取該latest_conversation_message_id爲每個用戶通話

下面的查詢獲取詳細信息user_id = 2，您可以修改，users.id = 2得到它的任何其他用戶

SQLFiddle，希望這能解決你的目的

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：查詢得到與任何其他用戶每次談話的最新消息，如果user_id 2，將消息發送到user_id 3，那也顯示，作爲它表示對話的開始。顯示與任何其他用戶的每次對話的最新消息

Answer 2

您的WHERE子句

ORDER BY "ColumnName" [ASC, DESC] 

UNION Description at W3Schools它結合了這2個語句的結果後粘貼。

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

Answer 3

要解決pg中的groupwise-max，可以使用DISTINCT ON。就像這樣：

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

要獲取最新的X但是我們將有子查詢中使用它：

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

要獲取發出和收到的線程：

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

對於大型數據集我想你可能想嘗試運行兩個語句，然後鞏固成果，作爲（USER_ID和id）或（recipient_id和id）索引掃描應該是非常有效的獲得每種類型的最近10次會話。

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 

編輯：其實另一個查詢值得嘗試的可能是：

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 

這可能是9.2+，其中單獨的指標可以被用來獲取ID的更好，或者在情況下，最最近檢索的數量非常大。儘管如此，還是有點不清楚。如果有疑問，我會去爲前一個版本。