从每个对话中获取最后一条消息

Question

我知道以前有人问过类似的问题，但他们都没有同样的条件，他们的答案不适用于这种情况。

包含消息的表格如下所示：

id | owner_id | recipient_id | content      | created
 1 |        1 |            2 | Hello        | 2015-12-08 20:00
 2 |        2 |            1 | Hey          | 2015-12-08 20:10
 3 |        3 |            1 | You there?   | 2015-12-08 21:00
 4 |        1 |            3 | Yes          | 2015-12-08 21:15
 5 |        4 |            1 | Hey buddy    | 2015-12-08 22:00

假设我从用户 ID 1 的每个对话中查询最后一条消息，预期结果是：

id | owner_id | recipient_id | content      | created
 5 |        4 |            1 | Hey buddy    | 2015-12-08 22:00
 4 |        1 |            3 | Yes          | 2015-12-08 21:15
 2 |        2 |            1 | Hey          | 2015-12-08 20:10

我尝试了很多组合，使用了 JOIN 和子查询，但没有一个给出预期的结果。

这是我尝试过的查询之一，但它不起作用。我相信它甚至不接近我需要的东西。

SELECT
    IF ( owner_id = 1, recipient_id, owner_id ) AS Recipient,

    (
        SELECT
            content
        FROM
            messages

        WHERE
            ( owner_id = 1 AND recipient_id = Recipient  )
        OR
            ( owner_id = Recipient AND recipient_id = 1 )

        ORDER BY
            created DESC

        LIMIT 1
    )
FROM
    messages

WHERE
    owner_id = 1
OR
    recipient_id = 1

GROUP BY
    Recipient;

Answer 1

select t.* 
    from 
        t 
      join 
        (select user, max(created) m  
            from 
               (
                 (select id, recipient_id user, created 
                   from t 
                   where owner_id=1 ) 
               union 
                 (select id, owner_id user, created
                   from t 
                   where recipient_id=1)
                ) t1
           group by user) t2
     on ((owner_id=1 and recipient_id=user) or 
         (owner_id=user and recipient_id=1)) and 
         (created = m)
   order by created desc

example on sqlfiddle

Answer 2

这应该可以解决问题：

$joins = array(
    array('table' => 'conversations',
        'alias' => 'Conversation2',
        'type' => 'LEFT',
        'conditions' => array(
            'Conversation.id < Conversation2.id',
            'Conversation.owner_id = Conversation2.owner_id',
        )
    ),
    array('table' => 'conversations',
        'alias' => 'Conversation3',
        'type' => 'LEFT',
        'conditions' => array(
            'Conversation.id < Conversation3.id',
            'Conversation.recepient_id = Conversation3.recepient_id',
        )
    )

);

$conditions = array(
    'OR' => array(
        array(
            'Conversation2.id'=>null,
            'Conversation.owner_id' => $ownerId
        ),
        array(
            'Conversation3.id'=>null,
            'Conversation.recipient_id' => $ownerId
        ),
     )
);

$order = array('Conversation.created'=>'DESC');

$lastConversations=$this->Conversation->find('all',compact('conditions','joins','order'));

假设表的名称是conversations，而您的模型名称是Conversation。它基于Retrieving the last record in each group 的公认答案中描述的技术。

Answer 3

这个解决方案最适合我。

    SELECT t1.*
    FROM chats AS t1
    INNER JOIN
    (
        SELECT
            LEAST(sender_id, receiver_id) AS sender_id,
            GREATEST(sender_id, receiver_id) AS receiver_id,
            MAX(id) AS max_id
        FROM chats
        GROUP BY
            LEAST(sender_id, receiver_id),
            GREATEST(sender_id, receiver_id)
    ) AS t2
        ON LEAST(t1.sender_id, t1.receiver_id) = t2.sender_id AND
           GREATEST(t1.sender_id, t1.receiver_id) = t2.receiver_id AND
           t1.id = t2.max_id
        WHERE t1.sender_id = ? OR t1.receiver_id = ?

Source