in where clause is ambiguous ต้องการกำหนดชื่อ db ในโค๊ดครับแต่ผมไม่ค่อยมีความรู้พอดีได้งานแก้ไขของอีกคนมา พอดีชื่อตัวแปลชนกันมันไม่ทราบว่าจะดึกจาก table ไหน

<?php
class cCompile{
	function __construct($db){
		$this->db = $db;
	}
	public function Login(){
		$result = array('status'=>'FAIL');
		
		$where = array(
			'agent_Code'			=> $_POST['agent'], บรรทัดนี้ครับ
			'staff_Username'	=> $_POST['username'], 
			'staff_Password'		=> $_POST['password']
		);
		$tbl = DB_PREFIX.'staff INNER JOIN '.DB_PREFIX.'agent ON '.DB_PREFIX.'agent.agent_Code = '.DB_PREFIX.'staff.agent_Code';
		$row = $this->db->record($tbl, $where);

		$where = array('staff_No' => $row->agent_LoggedInBy);
		$tbl = DB_PREFIX.'staff INNER JOIN '.DB_PREFIX.'agent ON '.DB_PREFIX.'staff.staff_No = '.DB_PREFIX.'agent.agent_LoggedInBy';
		$row_agent  = $this->db->record($tbl,$where);
			
		if($row === FALSE){
			$result['status'] = 'FAIL';
		}else{
			$result['status'] = 'COMPLETE';
			$this->db->update(DB_PREFIX.'staff', array('staff_LastLogin' => date("Y-m-d H:m:s")), array('staff_No' => $row->staff_No));
			$this->db->update(DB_PREFIX.'agent', array('agent_LastLoggedIn' => date("Y-m-d H:m:s"),'agent_LoggedInBy' => $row->staff_No), array('agent_No' => $row->agent_No));
			$_SESSION['LOGIN']['ID']			= $row->staff_No;
			$_SESSION['LOGIN']['STAFF_TYPE']	= $row->staff_Type;
			$_SESSION['LOGIN']['AGENT_CODE'] 	= $row->agent_Code;
			$_SESSION['LOGIN']['LAST']			= $row->agent_LastLoggedIn;
			$_SESSION['LOGIN']['LASTBY']		= $row_agent->staff_Fname.' '.$row_agent->staff_Lname;
			$_SESSION['LOGIN']['USER']			= $row->staff_Fname;
			$_SESSION['LOGIN']['ONLINE']		= CURRENT_TIME;
			//$_SESSION['LOGIN']['MENU'] 		= $row->admin_menu;
		}
		echo json_encode($result); 
	}
}
?>