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有类似的问题,但是他的问题是传递ID。但我的是别的东西。喜欢这个。但我得到错误。传递变量以在codeigniter中加入查询
$this->db->join($user_role, $user_role.'school_id = user.school_id', 'left');
给人Unknown column 'adminschool_id' in 'on clause'
$this->db->join($user_role, $user_role'.school_id = user.school_id', 'left');
给人unexpected ''.school_id = user.school_id''
什么是变量$ USER_ROLE价值? – ranakrunal9
它包含管理员,学生,老师。 – iamdevlinph
一次一个值,对吗?当你得到错误:意外''.school_id = user.school_id'',这意味着在那时$ user_role变量值是空的,因为它给你错误。你也可以尝试通过传递第四个参数给你的连接如下: $ this-> db-> join($ user_role,$ user_role'.school_id = user.school_id','left',FALSE); – ranakrunal9