Question

我试图在两个用户之间传递第一个信息,无论我是否发出这一信息。我提出以下问题,即回归所有正确之处,但似乎忽视了DISTINCT条款——我最后得出了相同结果。

我如何纠正这一疑问,以便不会重复在由此产生的 some栏中退还的价值?

www.un.org/Depts/DGACM/index_spanish.htm 这里指的是目前使用的Im:。

SELECT DISTINCT CASE WHEN $userid != senderid THEN senderid ELSE GROUP_CONCAT(receivers.id SEPARATOR  ,  ) END someid,   
            CASE WHEN $userid != senderid THEN senders.username ELSE GROUP_CONCAT(receivers.username SEPARATOR  ,  ) END somename,
            messages.body, 
            messages.time
    FROM messages 
    LEFT JOIN messages_recipients AS recipients ON messages.id = recipients.messageid
    LEFT JOIN users AS senders ON messages.senderid = senders.id
    LEFT JOIN users AS receivers ON recipients.userid = receivers.id
    WHERE recipients.userid = $userid
    OR messages.senderid = $userid
    GROUP BY messages.id
    ORDER BY messages.time DESC

Answer 1

如“Mitch Wheat”所述,你可能不得不将“DISTINCT”分为外部询问。

SELECT DISTINCT * FROM 
    (
    SELECT CASE WHEN ... GROUP BY messages.id ORDER BY messages.time DESC
    ) AS inner_q
ORDER BY messages.time DESC

但记住的是,DISTINCT不是单一价值(有些)。时间可能会妨碍你了解你之后的情况。

友情链接