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我有第三个表的外键标识,例如(具有的thirdsubmenu
表)。我想从我父母的表mainmenu
获得menu_name
的名字。 请参考下面我的数据库表结构完整的细节Codeigniter:无法使用连接获取字段值
数据库结构
1)Table: mainmenu
---------------
mainmenu_id PK(primary key)
menu_name .....
2)Table: submenu
-------------------
submenu_id PK
mainmenu_id FK (foreign key refrences mainmenu table)
submenu_name .....
3)Table: thirdsubmenu
--------------------
thirdsubmenu_id PK
submenu_id FK (foreign key refrences submenu table)
thirdsubmenu_name ........
我想从我的mainmenu
表中获取menu_name
下面的代码,但我得到的错误。我越来越
//---------------------------get Main Menu Name by thirdsubmenu_id-----------------------------------
function getMainMenuNameOfSubmenu($thirdsubmenu_id)
{
$this->load->database();
$this->db->select('*');
$query=$this->db->join('mainmenu', 'mainmenu.mainmenu_id = submenu.mainmenu_id', 'left')
->join('submenu', 'submenu.submenu_id = thirdsubmenu.submenu_id', 'left')
->get_where('thirdsubmenu',array('thirdsubmenu_id'=>$thirdsubmenu_id));
return $query->row('menu_name');
}
错误是:
A Database Error Occurred
Error Number: 1054
Unknown column 'submenu.mainmenu_id' in 'on clause'
SELECT * FROM (`thirdsubmenu`) LEFT JOIN `mainmenu` ON `mainmenu`.`mainmenu_id` = `submenu`.`mainmenu_id` LEFT JOIN `submenu` ON `submenu`.`submenu_id` = `thirdsubmenu`.`submenu_id` WHERE `thirdsubmenu_id` = '17'
Filename: D:\xampp\htdocs\system\database\DB_driver.php
Line Number: 330
我想你的代码。但得到错误'不是唯一的表/别名:“thirdsubmenu” ......' –
编辑where子句。再检查一遍。 –
对不起,编辑后,仍然有错误:'语法错误'在这一行附近' - > where('thirdsubmenu.thirdsubmenu_id =''。$ thirdsubmenu_id。''''));' –