php - 如何将从其他函数返回的变量与 mysql 中的匹配行进行比较

Question

我正在尝试比较 pr_resignation_requests 中作为 managerid 列存在的变量 $manager_id。如果存在则返回该行，否则不返回。但不知怎的，这个查询不起作用。尝试了很多东西但不起作用。我知道我的 where 子句有错误，

错误是:

Error Number: 1054

Unknown column '1' in 'where clause'

SELECT g.*、userids、resignations_date、reason_type、requested_date、last_status、date_last_status、approved_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 WHERE1= 'managerid'

我的查询是:

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;
}

Answer 1

您的查询有误。在where子句中第一个参数是表名而不是值

$query = $this->db->where('managerid', $manager_id);