这是我的查询运行良好,但想在其中添加另一个条件,但它给出了错误非唯一表/别名:'users'
这是我的代码
$query = 'select group_concat(DISTINCT evaluation_section. section_name) as allowed_sections from evaluation_section
join section_permissions on (section_permissions.section_id = evaluation_section.section_id )
join users on ( users.user_id = section_permissions.user_id )
join users on ( users.user_id = section_permissions.assigned_for )
where users.user_id = '.$user['user_id'];
也想从用户表中选择名称,然后使用此条件加入(users.user_id = section_permissions.assigned_for )
上的用户
请帮我
这是我编辑过的文本
这是我的完整工作查询
$q = 'select group_concat(DISTINCT evaluation_section. section_name) as allowed_sections from evaluation_section
join section_permissions on (section_permissions.section_id = evaluation_section.section_id )
join users on ( users.user_id = section_permissions.user_id )
where users.user_id = '.$user['user_id'];
但我希望它从users表中选择name,然后将其与users.user_id = section_permissions.assigned_for
匹配,这样它就可以获得assigned_for id name的名称和之前的group_concat一样,我使用users.name,然后它工作,但它显示users.user_id = section_permissions.user_id
的名称,但我想显示er_id的名称)其中users.user_id='$用户['user_id'];
但我希望它从用户表中选择名称,然后将其与users.user_id = section_permissions.assigned_for
匹配在同一个查询中,因为在两个中它不工作
试着这样做:
$query = 'select group_concat(DISTINCT evaluation_section. section_name) as allowed_sections from evaluation_section
join section_permissions on (section_permissions.section_id = evaluation_section.section_id )
join users u1 on ( u1.user_id = section_permissions.user_id )
join users u2 on ( u2.user_id = section_permissions.assigned_for )
where u1.user_id = '.$user['user_id'];
由于要加入"用户"表两次,所以必须为这两个用户提供唯一的别名。