像codeigniter中的从句一样


like clause in codeigniter

我想在last_name的情况下包含like子句。这是我的控制器代码用于搜索框。我想比较表单值last_name与数据库值family_name。

public function filtered_volunteer_details()
{
    $user_data = $this->session->userdata("logged_user");
    $tab_result = array();
    if (is_array($user_data) && $user_data["user_role"] == "staff")
    {
        if ($this->input->method() == "post")
        {
            $condition = array();
            $data = $this->input->post();
            if(isset($data["membership_number"]) && $data["membership_number"]!= "")
            {
                $condition["membership_number"]=trim($data["membership_number"]);
            }
            if(isset($data["last_name"]) && $data["last_name"]!= "")
            {
                $condition["family_name "]=trim($data["last_name"]);
            }
            $this->load->model(array(
                "Users"));
            $result = $this->Users->get_volunteers("*",$condition);
            $volunteer_tbl_list=array();
            foreach($result as $volunteer_results)
            {
            $volunteer_tbl_list[] = $this->load->view("staff/manage_volunteer/volunteer_list_tbl_row",  $volunteer_results, TRUE);  
            }
            echo json_encode(array(
            "status" => "success",
            "volunteer_tbl_list" => $volunteer_tbl_list));
            return;
        }
    }
    echo json_encode(array(
        "error" => "true",
        "msg" => "Invalid Access."));
    return;
}

这是我的模型代码

public function get_volunteers($fields = "*",$condition=array())
{
    $condition["staff"]="N";
    $this->db->select($fields);
    $query = $this->db->get_where(self::$tbl_name, $condition);
    //var_dump($this->db->last_query());
    return( $query->result() );
}

尝试将$this->db->like()添加到get_volunteers()的查询中。文档

public function get_volunteers($fields = "*",$condition=array())
{
    $condition["staff"]="N";
    $this->db->select($fields);
    $this->db->like($condition);
    $this->db->from(self::$tbl_name);
    return( $query->result() );
}