Question

我正在使用Codeigniter编写一个小应用程序，到目前为止遇到了一些问题！我有一个从db表填充的下拉列表。下拉列表已填充并正确显示，但是当我尝试将其内容（值）保存到新的db表时，它将保存下拉列表中的位置，而不是显示的值。

视图中的部分表单：

    <?php echo form_open('manage/create_sold_vehicle_record'); ?>
    Registration Number
    <br>
    <?php echo form_dropdown('reg_no', $vehicle_list, 'large'); ?>

控制器的相关部分：

    function sell_vehicle()
    {
    $this->load->helper('form');
    $this->load->helper('url');

    $id = $this->uri->segment(3);
    $this->load->model('MVehicle','',TRUE);
    $this->load->model('manage_model','',TRUE);
    $vehicles_qry = $this->MVehicle->listSaleVehicles();

    $vehicle_list = array();
    foreach ($vehicles_qry->result() as $vehicle)
    {
        $vehicle_list[] = $vehicle->reg_no;
    }

    $data['vehicle_list'] = $vehicle_list;
    $data['main_content'] = 'sell_stock_vehicle';
    $this->load->view('includes/template', $data);  
    }

这是模特：

    function create_sold_vehicle_record()
    {


        $new_vehicle_insert_data = array(
            'reg_no' => $this->input->post('reg_no'), <-- This is saved to db, but is the position in dropdown list, not the value shown.
            'sold_by' => $this->input->post('staff'),
            'bought_by' => $this->input->post('customer'),
            'sale_price' => $this->input->post('sale_price'),
            'comm_rate' => $this->input->post('comm_rate'),
            'total_comm' => $this->input->post('total_comm'),
            'sale_date' => $this->input->post('date')
        );

        $insert = $this->db->insert('sold_vehicles', $new_vehicle_insert_data);

        return $insert;
    }

所以，而不是例如保存＆＃39; OU03XHC＆＃39;作为db中的reg_no值，它正在保存＆＃39; 1＆＃39; （或者值在下拉列表中的位置）。

Answer 1

$vehicle_list[]需要是一个关联数组

$vehicle_list = array();
foreach ($vehicles_qry->result() as $vehicle)
{
    $vehicle_list[$vehicle->reg_no] = $vehicle->reg_no;
}

现在，form_dropdown（）还会在下拉列表中生成选项值（0,1等除外）

Answer 2

替换

$vehicle_list = array();
 foreach ($vehicles_qry->result() as $vehicle)
 {
   $vehicle_list[] = $vehicle->reg_no;
 }

TO

$vehicle_list = array();
  foreach ($vehicles_qry->result() as $vehicle)
  {
      $vehicle_list[$vehicle->reg_value] = $vehicle->reg_no;
  }

“$ vehicle-＆gt; reg_value”veriable是您要存储在数据库中的值

CodeIgniter - 下拉菜单将列表中的位置传递给db而不是value

2 个答案: