如何将从其他函数返回的变量与mysql中的匹配行进行比较

我正在尝试比较变量$manager_id作为managerid列存在于pr_resignation_requests中。如果有，则返回行，否则不返回。但不知怎么的，这个查询不起作用。尝试了很多东西，但都不起作用。我知道我的where子句有错误，

错误为：

错误编号：1054

"where子句"中的未知列"1"

SELECT g.*, userids, resignations_date, reason_type, requested_date, last_status, date_last_status, agreed_date, exit_details, exit_checklist, firstname, lastname, managerid FROM (pr_resignation_requests as g) JOIN pr_users_details as ud ON ud.userid = g.userids WHERE 1 = 'managerid'

我的问题是：

function get_resignation_request($id=0)
{
global $USER;
$post_arr = $this->input->post();
$manager_id = $this->get_value_by_id('managerid','users',$this->session->userdata('admin_id'));
$this->db->select('g.*,userids,resignations_date,reason_type,requested_date,last_status,date_last_status,agreed_date,exit_details,exit_checklist,firstname,lastname,managerid');
$this->db->from('pr_resignation_requests as g'); 
$this->db->where($manager_id, managerid); 
//$where = "$manager_id='1'";
	//$this->db->where($where);
//$this->db->join('pr_resignation_requests as uds','uds.managerid = ".$manager_id" ');
//$this->db->where($manager_id, managerid); 
//$this->db->where($manager_id = managerid); 
//$this->db->join($this->myTables['pr_users_details'].' as ud','ud.userid = g.userid');
$this->db->join('pr_users_details as ud','ud.userid = g.userids');
//$this->db->join('pr_users as uds','uds.id = g.managerid');
/*$this->db->join('pr_resignation_type as gt','gt.id = g.sr_type');*/
$query=$this->db->get();	
$return	= $query->result_array();
return $return;
}