我有下表：
代理
+-------------------------------------+
| id | name | desc | comment |
|-------------------------------------+
| 1 | agent1 | agent 1 | sss |
| 2 | agent2 | agent 2 | eee |
|-------------------------------------|
agent_old
+-------------------------------------+
| id | name | desc | comment |
|-------------------------------------+
| 1 | agent1 | agent 1 | sss |
| 2 | agent3 | agent 3 | eee |
|-------------------------------------|
身份验证
+-------------------------------+
| id | login | password |
|-------------------------------+
| 1 | agent1 | xxxxxxx |
| 2 | agent2 | yyyyyy |
|-------------------------------|
auth_old
+-------------------------------+
| id | login | password |
|-------------------------------+
| 1 | oldagent1 | wwwwww |
| 2 | oldagent2 | qqqqqq |
|-------------------------------|
我需要这样的结果表：
代理
+-------------------------------------+
| id | name | desc | comment |
|-------------------------------------+
| 1 | agent1 | agent 1 | sss |
| 2 | agent2 | agent 2 | eee |
|-------------------------------------|
身份验证
+-------------------------------+
| id | login | password |
|-------------------------------+
| 1 |oldagent1 | wwwwww |
| 2 | agent2 | yyyyyy |
|-------------------------------|
这是我得到的但没有运行：
update auth a
set
a.login = oa.login,
a.password = oa.password
from (
select o.login,
o.password
from auth_old o
join agent na
join agent_old ago
on ago.id = o.id
and na.name = ago.name
and na.desc = ago.desc
) oa
where a.id = na.id
答案 0 :(得分：1)
在mysql中，您可以使用此sintax，但您在（select）oa中没有id。我为此添加了o.id（希望是正确的）
update auth a
inner join (
select o.login,
o.password ,
na.id
from auth_old o
join agent na
join agent_old ago
on ago.id = o.id
and na.name = ago.name
and na.desc = ago.desc
) oa on a.id = oa.id
set
a.login = oa.login,
a.password = oa.password
（并且按照Bill Karvin的建议，您使用了错误的表别名na而不是oa）