Question

我有一个表单，我用它来允许人们输入数据然后上传预设图片。我想将该图片标记为form_id。我可以运行LAST_INSERT_ID（）查询并在我在视图中插入表单后显示它，但我似乎无法在其他任何地方回显该信息。我需要在$ update查询后运行select_last_id的查询，否则ID号将关闭。任何人都可以帮助我将行的值传递给我的视图吗？这是我的控制器的代码。

function inspection() {

    if($this->input->post('submit')) {
        $array = array(

            'trlr_num' => $this->input->post('trlr_num'),
            'seal' => $this->input->post('seal'),
            'damaged' => $this->input->post('damaged'),
            'truck_num' => $this->input->post('truck_num'),
            'driver_name' => $this->input->post('driver_name'),
            'car_code' => $this->input->post('car_code'),
            'origin' => $this->input->post('origin'),
            'lic_plate' => $this->input->post('lic_plate'),
            'del_note' => $this->input->post('del_note'),
            'live_drop' => $this->input->post('live_drop'),
            'temp' => $this->input->post('temp'),
            'level' => $this->input->post('level'),
            'ship_num' => $this->input->post('ship_num'),
            'trlr_stat' => $this->input->post('trlr_stat'),
            'comment' => $this->input->post('comment')

                        );
        $update = $this->trailer_model->insert_form($array);

        $query = $this->trailer_model->select_last_id();
        $result =& $query->result_array();

        $this->table->set_heading('ID');
        $data['table'] = $this->table->generate_table($result);
        unset($query,$result);


       }
    $level = $this->trailer_model->select_fuel_level();
    $result = $level->result_array();

    $data['options'] = array();
    foreach($result as $key => $row) {
        $data['options'][$row['level']] = $row['level'];
    }
    unset($query,$result,$key,$row);

    $data['label_display'] = 'Fuel Level';
    $data['field_name'] = 'level';

    $status = $this->trailer_model->select_trailer_type();
    $result = $status->result_array();

    $data['options1'] = array();
    foreach($result as $key => $row) {
        $data['options1'][$row['trlr_stat']] = $row['trlr_stat'];
    }
    unset($query,$result,$key,$row);


    $data['label_display1'] = 'Trailer Status';
    $data['field_name1'] = 'trlr_stat';


    $data['page_title'] = 'Trailer Inspection';
    $data['main_content'] = 'dlx/inspection/trailer_inspection_view';


    return $this->load->view('includes/template',$data);

}

}

Answer 1

您要在视图上显示的任何内容，都必须传递给它。

当您设置$data时，您错过了向其添加$query。

<强>控制器：

我知道$query = $this->trailer_model->select_last_id();只返回一个ID，所以：

$data['last_id'] = $query;

但如果$query = $this->trailer_model->select_last_id(); 不只返回一个ID，而是一个数组或其他内容，则应包括在模型方法中返回ID。实际上，了解$this->trailer_model->select_last_id();正在返回的内容至关重要。

查看：

echo $last_id; - ＆gt;如果模型中的select_last_id()方法返回它必须返回的内容，则必须回显最后一个ID。

无法将SQL数据传递给codeigniter中的视图

1 个答案: