mysql - 从3个表中选择数据

Question

我正在尝试从数据库中获取一些数据。

1. 用户名 来自登录表
2. email 来自联系人表

检查两个表 tutor 和 institute 中的 2 个值

到目前为止，这是我的代码:

SELECT s. * , c.email, l.username
FROM (
        SELECT  contact_id AS id, 
                  login_id, 
                  username, 
                  tutor_code AS code, 
                  tutor_name AS Name, 
                  'tutor' AS profile
        FROM tutors
        WHERE tutor_code = $code AND tutor_name = '$name'
        UNION ALL
        SELECT  contact_id AS id, 
                  login_id, 
                  username, 
                  institute_code AS code, 
                  institute_name AS Name, 
                  'institute' AS profile
        FROM institutes
        WHERE institute_code = $code AND institute_name = '$name'
        )
INNER JOIN contact c ON s.id = c.contact_id
INNER JOIN login l ON s.login_id = l.login_id

此查询无效，并且有一条错误消息。

1054 - Unknown column 'username' in 'field list'

更新

SELECT s. * , c.email, l.username
FROM (
        SELECT  contact_id AS id, 
                  login_id, 
                  username, 
                  tutor_code AS code, 
                  tutor_name AS Name, 
                  'tutor' AS profile
        FROM tutors
        WHERE tutor_code = $code AND tutor_name = '$name'
        UNION ALL
        SELECT  contact_id AS id, 
                  login_id, 
                  username, 
                  institute_code AS code, 
                  institute_name AS Name, 
                  'institute' AS profile
        FROM institutes
        WHERE institute_code = $code AND institute_name = '$name'
        )s
INNER JOIN contact c ON s.id = c.contact_id
INNER JOIN login l ON s.login_id = l.login_id

Answer 1

由于您正在从 login 中检索您的 username，username 列很可能不存在于 tutors 和/或 institutes，加入 login 也不是必需的，因为您是通过 login_id 加入的，我想您可以只需从子查询中删除 username 列:

SELECT s. * , c.email, l.username
FROM (
        SELECT  contact_id AS id, 
                  login_id, 
                  --username, 
                  tutor_code AS code, 
                  tutor_name AS Name, 
                  'tutor' AS profile
        FROM tutors
        WHERE tutor_code = $code AND tutor_name = '$name'
        UNION ALL
        SELECT  contact_id AS id, 
                  login_id, 
                  --username, 
                  institute_code AS code, 
                  institute_name AS Name, 
                  'institute' AS profile
        FROM institutes
        WHERE institute_code = $code AND institute_name = '$name'
        ) s
INNER JOIN contact c ON s.id = c.contact_id
INNER JOIN login l ON s.login_id = l.login_id

我还向您的 subuqery 添加了别名 s，因为我认为它的遗漏是一个打字错误，因为如果没有它会引发语法错误