[eluser]ashutosh[/eluser]
Currently i have foreign key thirdsmcontent_id value , now i want to get the value of
`thirdsubmenu_name` from `thirdmenu` table,`submenu_name` from `submenu` table,`menu_name` from `mainmenu` table .
Please help me to solve this problem.
I tried below code but my model code is not correct. it is not display my data
Code:
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 ........
4)Table: thirdsmcontentdetails
--------------------
thirdsmcontent_id PK
thirdsubmenu_id FK (foreign key refrences thirdsubmenu table)
content ......
In My controller
Code:
$thirdsubmenu_id = $this->uri->segment(4);
$data['main_menuname'] = $this->thirdsmcontentdetailsmodel->getMainMenuNameOfSubmenu($thirdsubmenu_id);
$data['sub_main_menuname'] = $this->thirdsmcontentdetailsmodel->getSubMenuNameSubmenu($thirdsubmenu_id);
$data['third_sub_main_menuname'] = $this->thirdsmcontentdetailsmodel->getThirdSubMenuNameSubmenu($thirdsubmenu_id);
In My model
Code:
//---------------------------get Main Menu Name by Menu id-----------------------------------
function getMainMenuNameOfSubmenu($thirdsubmenu_id)
{
$this->load->database();
$query = $this->db->join('thirdsubmenu','thirdsubmenu.submenu_id = thirdsmcontentdetails.submenu_id')->get_where('thirdsubmenu',array('thirdsubmenu_id'=>$thirdsubmenu_id));
return $query->row('menu_name');
}
//---------------------------get Sub Menu Name by Menu id-----------------------------------
function getSubMenuNameSubmenu($thirdsubmenu_id)
{
$this->load->database();
$query = $this->db->join('mainmenu', 'mainmenu.mainmenu_id = submenu.mainmenu_id')->get_where('thirdsubmenu',array('thirdsubmenu_id'=>$thirdsubmenu_id));
return $query->row('submenu_name');
}
//---------------------------Get the third menu name-------------------------------------
function getThirdSubMenuNameSubmenu($thirdsubmenu_id)
{
$this->load->database();
$query = $this->db->get_where('thirdsubmenu',array('thirdsubmenu_id'=>$thirdsubmenu_id));
return $query->row('thirdsubmenu_name');
}
Getting Error:
Code:
A Database Error Occurred
Error Number: 1066
Not unique table/alias: 'thirdsubmenu'
SELECT * FROM (`thirdsubmenu`) JOIN `thirdsubmenu` ON `thirdsubmenu`.`submenu_id` = `thirdsmcontentdetails`.`submenu_id` WHERE `thirdsubmenu_id` = '1'
Filename: D:\xampp\htdocs\system\database\DB_driver.php
Line Number: 330
