php - 将用户 ID 传递给嵌套查询

Question

我试图从用户数据库中选择未订阅特定邮件列表的所有用户。如果用户是列表的成员，则内部查询会查找该用户，而外部查询会选择内部结果为零行的所有用户。

SELECT u.userid, u.username, u.mail, u.name, 
    u.lastname, u.starting_year, u.userid AS UIDD 
FROM userdb.users AS u 
WHERE (SELECT COUNT(*) FROM maildb.lists AS l, 
        maildb.list_subscriptions AS s, userdb.users AS u 
    WHERE u.userid = UIDD 
    AND l.listid = '$LIST_ID_TO_LOOKUP' 
    AND s.userid = u.userid 
    AND s.listid = l.listid) = 0 
ORDER BY u.starting_year DESC, u.lastname, u.name;

这一切都很好，直到我搬到新服务器并更新了许多组件，包括 MySQL。

现在运行此查询时出现错误:

#1054 - Unknown column 'UIDD' in 'where clause'

这可能是现在已弃用的旧语法的问题吗？我现在运行的是 MySQL 5.5.24。

Answer 1

像这样，SELECT .. WHERE userid NOT IN(子查询)

SELECT 
 u.userid, 
 u.username, 
 u.mail, 
 u.name, 
 u.lastname,
 u.starting_year, 
 u.userid AS UIDD 
FROM 
 userdb.users AS u 
WHERE 
 u.userid NOT IN (
    SELECT 
      u.userid 
    FROM 
      maildb.lists AS l, 
      maildb.list_subscriptions AS s, 
      userdb.users AS u 
    WHERE 
      u.userid = u.userid 
      AND l.listid = '$LIST_ID_TO_LOOKUP' 
      AND s.userid = u.userid 
      AND s.listid = l.listid
    ) 
ORDER BY 
 u.starting_year DESC, u.lastname, u.name;