mysql - 选择union中的最大值

Question

给出两个这样的表：

用户

+--------+-----------+
| userid | username  |
+--------+-----------+
|      1 | username1 |
|      2 | username2 |
|      3 | username3 |
|      4 | username4 |
|      5 | username5 |
+--------+-----------+

会议

+--------+-----------+
| userid | timestamp |
+--------+-----------+
|      1 |       123 |
|      3 |       321 |
|      5 |       135 |
|      1 |       222 |
|      5 |       333 |
+--------+-----------+

我想找到每个用户的最大会话时间戳。 这看起来很好：

select users.username, max(s.timestamp) as timestamp from sessions s
    left join users on s.userid=users.userid group by s.userid
    order by timestamp;

+-----------+-----------+
| username  | timestamp |
+-----------+-----------+
| username1 |       222 |
| username3 |       321 |
| username5 |       333 |
+-----------+-----------+

对于没有会话条目的所有用户，我希望看到时间戳为0 ，所以我试试这个：

select users.username, coalesce(max(s.timestamp), 0) as timestamp from sessions s
    left join users on s.userid=users.userid
union
select users.username, coalesce(max(s.timestamp), 0) as timestamp from sessions s
    right join users on s.userid=users.userid
group by users.userid order by timestamp;

结果如下：

+-----------+-----------+
| username  | timestamp |
+-----------+-----------+
| username2 |         0 |
| username4 |         0 |
| username1 |       222 |
| username3 |       321 |
| username5 |       333 |
| username1 |       333 |
+-----------+-----------+

它看起来几乎不错（用户2和4按预期得到0），但“username1”与实际属于“username5”的会话中的值重复。

是什么导致这种情况以及如何解决？

Answer 1

left join是正确的方法。 只是users应该是第一个表：

select u.username, coalesce(max(s.timestamp), 0) as timestamp
from users u left join
     sessions s
     on s.userid = u.userid
group by u.username
order by timestamp;

请注意，我还按键更改了group by以匹配select列。

Answer 2

反转主表（将用户作为主表而不是sessione）并检查是否为null
你可以用

select users.username
 , case when max(s.timestamp) is null then 0 else max(s.timestamp) end  as timestamp 
from users 
left join session  s on s.userid=users.userid
group by user.username
order by timestamp;