我有两个视图，维修和维护。这两个视图都是三个表的组合：维修、r_列表、物料;我需要将repair的id和maintenance的id放入第三个表tickets中名为ticket_id的一列中。所以基本上应该是这样的：three tables plus three tables into one table combined
这三个表必须同时存在，这意味着维修中的任何更改都将影响到票证，维护中的情况也是如此。

型号：

public function get_tickets_view()
        {
            $this->db->select("
                tickets.id AS id,

                repair.repair_id AS ticket_id,
                maintenance.maintenance_id AS ticket_id,

                items.item_id AS item_id,

            ");

            $this->db->from("tickets");
            $this->db->join("items", "tickets.item_id = items.item_id", "left");
            $this->db->join("repair", "repair.item_id = tickets.item_id", "left");
            $this->db->join("maintenance", "maintenance.item_id = tickets.item_id", "left");

            $query = $this->db->get();
            return $query->result();
        }

控制器

public function tickets()
        {
            $data['ticket'] = $this->sample_model->get_tickets_view();

            $data['main_content'] = 'page_address';
            $this->load->view('navbar', $data);

        }

尽管页面显示正确，但它实际上拒绝将项目拉入这两个表视图中的任何一个。