Question

我正在尝试在两个用户之间返回第一条消息，无论是我发送了消息还是收到了消息。我编写了以下查询，它返回所有正确的线程，但似乎忽略了DISTINCT子句 - 我最终得到的结果与'someid'相同。

如何更正此查询，以便在结果'someid'列中不会返回重复值？

以下是我目前正在使用的内容：

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

正如@Mitch Wheat所提到的，您可能必须将DISTINCT分离为外部查询。

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

但请记住，DISTINCT是针对不同的记录而不是单个值（someid）。 messages.time可能会阻止你获得你想要的东西。

SELECT DISTINCT在CASE语句后不起作用

1 个答案: