1054 - “字段列表”

Question

我正在尝试从db获取一些数据。

username
email来自联系表

使用两个表tutor和institute

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，并且出现错误消息。

1054 - “字段列表”
中的未知列'用户名'

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

由于您似乎从username检索了login，因此username和/或tutors中很可能不存在institutes列，加入login也没有必要因为您加入login_id，我认为您可以从子查询中删除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

我还在你的subuqery中添加了别名s，因为我认为它的遗漏是一个错字，因为它会在缺席时抛出语法错误

Answer 2

无需从username和tutors表调用institutes，并在子查询关闭时使用as abc

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

此查询将返回重复条目，因为您在ALL中使用union。

希望它适合你。

Answer 3

子查询中的username字段未找到，您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

从3个表中选择数据

1054 - “字段列表”

3 个答案: