这个MySQL查询有什么问题？SELECT*AS`x`,以后如何再次使用x？

Question

以下MySQL查询:

select `userID` as uID,
(select `siteID` from `users` where `userID` = uID) as `sID`,
from `actions`
where `sID` in (select `siteID` from `sites` where `foo` = "bar")
order by `timestamp` desc limit 100

...返回错误:

Unknown column 'sID' in 'IN/ALL/ANY subquery'

我不明白我在这里做错了什么.这个sID东西不应该是一个列,而是我通过执行创建的'别名'(这叫做什么？)(select siteID from users where userID = uID) as sID.它甚至不在IN子查询中.

有任何想法吗？

---

编辑: @Roland:感谢您的评论.我有三张桌子actions,users和sites.该表actions包含一个userID字段,该字段对应于表中的条目users.此表(users)中的每个用户都有一个siteID.我正在尝试从actions表中选择最新的操作,并将它们链接到users和sites表,以找出执行这些操作的人员以及在哪个站点上.希望有道理:)

Answer 1

您需要将其括在子查询中:

SELECT  *
FROM    (
        SELECT  userID as uID, (select siteID from users where userID = actions.userID) as sID,
        FROM    actions
        ) q
WHERE   sID IN (select siteID from sites where foo = "bar")
ORDER BY
        timestamp DESC
LIMIT   100

,或者,更好的是,将其重写为 JOIN

SELECT  a.userId, u.siteID
FROM    actions a
JOIN    users u
ON      u.userID = a.userID
WHERE   siteID IN
        (
        SELECT  siteID
        FROM    sites
        WHERE   foo = 'bar'
        )
ORDER BY
        timestamp DESC
LIMIT   100

创建以下索引:

actions (timestamp)
users (userId)
sites (foo, siteID)