MySql查询，用于从对话和消息表联接中选择最新消息

Question

请告诉我这段代码有什么问题，它返回所有对话，但仅返回第一次对话的最新消息。

    SELECT 
        conversations.*,
        messages.message,
        patients.first_name as fullname,
        patients.city,
        patients.thumb,
        patients.gender,
        patients.online_status
    FROM 
        patients
    INNER JOIN conversations
        ON conversations.patient_id_fk = patients.id
    LEFT JOIN messages
        ON messages.conversation_id = conversations.id
        AND messages.message_id = 
        (
           SELECT MAX(message_id) 
           FROM messages z
           WHERE z.therapist_id_fk = conversations.therapist_id_fk
        )
    WHERE conversations.therapist_id_fk='1'
    GROUP BY conversations.id
    ORDER BY messages.message_id DESC

Answer 1

您想要对话的最大信息而不是治疗师 ，所以我怀疑您希望在此子查询中这样做：

    AND messages.message_id = 
    (
       SELECT MAX(message_id) 
       FROM messages z
       WHERE z.message_id = conversations.id
    )

我不确定外部的GROUP BY是否正确（可能根本不需要它，并且似乎与SELECT的列冲突），但是如果没有示例数据和所需的结果，这很难分辨。

编辑：

我知道，问题在于外部查询中对治疗师ID的过滤。 这使它变得更加复杂：

    AND messages.message_id = 
    (
       SELECT MAX(message_id) 
       FROM messages z JOIN
            conversations c
            ON m.conversation_id = c.id
       WHERE c.therapist_id_fk = 1 AND  -- restriction on subquery
             z.message_id = conversations.id  -- correlation to outer query
    )

除非列是字符串或日期，否则不要对常量使用单引号。

Answer 2

    SELECT 
        conversations.*,
        messages.message,
        patients.first_name as fullname,
        patients.city,
        patients.thumb,
        patients.gender,
        patients.online_status
    FROM 
        patients
    INNER JOIN conversations
        ON conversations.patient_id_fk = patients.id
    LEFT JOIN messages
        ON messages.conversation_id = conversations.id
        AND messages.message_id = 
        (
           SELECT MAX(message_id) 
           FROM messages z
           WHERE z.conversation_id = conversations.id
        )
    WHERE conversations.therapist_id_fk='1'
    ORDER BY messages.message_id DESC

找到了问题。 在这里：

AND messages.message_id = 
            (
               SELECT MAX(message_id) 
               FROM messages z
               WHERE z.conversation_id = conversations.id
            )

我正在使用这个：

AND messages.message_id = 
        (
           SELECT MAX(message_id) 
           FROM messages z
           WHERE z.therapist_id_fk = conversations.therapist_id_fk
        )

Answer 3

我可以通过此查询解决问题

SELECT conversations.*,m.message FROM `conversations` LEFT JOIN messages as m on conversations.id = m.conversation_id AND m.id = (SELECT MAX(ms.id) from messages as ms where ms.conversation_id = conversations.id)

我离开了邮件，这样做时，我选择了最上面的邮件。 Order by可用于按日期给出订单

Answer 4

您的问题是：

SELECT MAX(message_id) 
FROM messages z
WHERE z.therapist_id_fk = conversations.therapist_id_fk

此查询仅匹配一条消息。 您需要一些不同的东西。 尝试这个：

SELECT 
    conversations.*,
    messages.message,
    patients.first_name as fullname,
    patients.city,
    patients.thumb,
    patients.gender,
    patients.online_status
FROM 
    patients
INNER JOIN conversations
    ON conversations.patient_id_fk = patients.id
LEFT JOIN messages
    ON messages.conversation_id = conversations.id        
WHERE conversations.therapist_id_fk='1' 
      AND messages.message_id IN (  -- new bit
          SELECT MAX(message_id) FROM messages z GROUP BY z.conversation_id
      )
GROUP BY conversations.id
ORDER BY messages.message_id DESC