从3个表中选择数据

Question

我想从数据库中获取一些数据。

1. username从接触表

登录表

email在两个表tutor和institute

与2个值检查这是到目前为止我的代码：

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 

此查询不起作用，并且有错误消息。

1054 - 在 '字段列表'

UPDATE

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

因为它似乎要检索您的username从login，列username最有可能不tutors和/或institutes存在，也没有必要，因为你是在login_id加盟加盟login ，我想你可以只从子查询中删除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 

我还添加别名s您subuqery因为我以为我T的遗漏是一个错字，因为它会在没有抛出一个语法错误

Answer 2

这是在子查询中username场未发现未知列 '用户名'，你必须在子查询中包含login表。

SELECT s. * , c.email, l.username 
FROM (
     SELECT contact_id AS id, 
        login_id, 
        l1.username, 
        tutor_code AS code, 
        tutor_name AS Name, 
        'tutor' AS profile 
     FROM tutors t 
     INNER JOIN login l1 ON l1.login_id = t.login_id 
     WHERE tutor_code = $code AND tutor_name = '$name' 
     UNION ALL 
     SELECT contact_id AS id, 
        login_id, 
        l2.username, 
        institute_code AS code, 
        institute_name AS Name, 
        'institute' AS profile 
     FROM institutes i 
     INNER JOIN login l2 ON l2.login_id = i.login_id 
     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 

Answer 3

没有必要从tutors和institutes表调用username和使用as abc当你的子查询关闭像

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

这个查询将返回重复条目，因为您在union中使用ALL。

希望它适合你。