我们可以尝试使用带有 EXISTS 条款：

SELECT DISTINCT mu1.convId
FROM messages_users mu1
INNER JOIN messages_users mu2
    ON mu1.convId = mu2.convId
WHERE
    (mu1.userId = 1 AND mu2.userId = 2) AND
    NOT EXISTS (SELECT 1 FROM messages_users mu
                WHERE mu.convId = mu1.convId AND userID NOT IN (1, 2));

下面是一个演示的链接，演示了上述逻辑的工作原理：

演示

您需要的条件是：

SELECT mu.convId
FROM messages_users mu
GROUP BY mu.convId
HAVING SUM(CASE WHEN mu.userId = 1 THEN 1 ELSE 0 END) > 0 AND
       SUM(CASE WHEN mu.userId = 2 THEN 1 ELSE 0 END) > 0 AND
       SUM(CASE WHEN mu.userId NOT IN (1, 2) THEN 1 ELSE 0 END) = 0;

你不能在 GROUP BY 因为如果在对话中，您将无法计算用户“3”。第三个条件阻止“3”参与。
在mysql中，这可以简化为：

SELECT mu.convId
FROM messages_users mu
GROUP BY mu.convId
HAVING SUM( mu.userId = 1 ) > 0 AND
       SUM( mu.userId = 2 ) > 0 AND
       SUM( mu.userId NOT IN (1, 2) ) = 0;

对于每个“与朋友的对话”，必须只有两行具有相同的 convId :

SELECT convId
FROM messages_users 
GROUP BY convId
HAVING COUNT(userId) = 2;

如果您想要用户1和2的“与朋友的对话”：

SELECT DISTINCT convId FROM messages_users 
WHERE 
  (userId IN (1, 2)) 
  AND 
  convId IN 
    (SELECT convId
    FROM messages_users 
    GROUP BY convId
    HAVING COUNT(userId) = 2);

请参见演示