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有類似的問題,但是他的問題是傳遞ID。但我的是別的東西。喜歡這個。但我得到錯誤。傳遞變量以在codeigniter中加入查詢
$this->db->join($user_role, $user_role.'school_id = user.school_id', 'left');
給人Unknown column 'adminschool_id' in 'on clause'
$this->db->join($user_role, $user_role'.school_id = user.school_id', 'left');
給人unexpected ''.school_id = user.school_id''
什麼是變量$ USER_ROLE價值? – ranakrunal9
它包含管理員,學生,老師。 – iamdevlinph
一次一個值,對嗎?當你得到錯誤:意外''.school_id = user.school_id'',這意味着在那時$ user_role變量值是空的,因爲它給你錯誤。你也可以嘗試通過傳遞第四個參數給你的連接如下: $ this-> db-> join($ user_role,$ user_role'.school_id = user.school_id','left',FALSE); – ranakrunal9