如何从其他表单中获取值

时间:2017-11-15 11:27:57

标签: php forms codeigniter web

现在我试图从另一种形式获取价值。但我不知道如何制作它。

这是我的第一个表格:

click to see

这是方法,当我们在form1中单击“提交”时,页面将重定向到此方法。

public function input_data_inserted()
{
    $id_data_inserted = $this->input->post('id_data_inserted');
    $date = $this->input->post('date');
    $id_data = $this->input->post('id_data');
    $amount= $this->input->post('amount');
    $id_room = $this->input->post('id_room');

    $passed = FALSE;

    $checkIdData = $this->modelku->cek_idData()->result_array();
    $checkIdRoom = $this->modelku->cek_idRoom()->result_array();

    foreach ($cekIdData as $cID) 
    {
        foreach ($cekIdRoom as $cIR) 
        {
            if ($id_data == $cID['id_data'] && $id_room == $cIR['id_room']) 
            {
                $data = array 
                (
                    'id_data_inserted' => $id_data_inserted,
                    'date' => $date,
                    'id_data' => $cID['id_data'],
                    'amount' => $amount,
                    'id_room' => $cIR['id_room'],
                );

                $this->modelku->input_data($data, 'table_data_inserted');
                $data['table_data_inserted'] = $this->modelku->show_data('table_data_inserted')->result();
                $this->modelku->increase_value($amount, $id_data);
                //redirect to second form
                redirect('admin/Data/data\view_detail_data');

                $passed = TRUE;
                break;
            }
        }
    }

    if(!$passed)
    {
        echo "ERORR";
    }
}   

这是我的第二种形式:

click to see

第二种形式的方法:

public function input_detail_barang()
{
    $id_data = $this->input->post('id_data');
    $no_inv = $this->input->post('no_inv');
    $condition = $this->input->post('condition');

    $data = array 
    (
        'id_data' => $id_data,
        'no_inv' => $no_inv,
        'condition' => $condition
    );

    //I need amount from first form, because i will use it. How can i do?


}

任何解决方案? THX

1 个答案:

答案 0 :(得分:0)

在此行之后,您可以将其保存到会话:

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table>
<tr class='group'> <td>Group 1 </td> </tr>
<tr> <td> Group 1 item 1</td> </tr>
<tr> <td> Group 1 item 2</td> </tr>
<tr> <td> Group 1 item 3</td> </tr>
<tr class='group'> <td> group 2</td> </tr>
<tr> <td> group 2 item 1</td> </tr>
<tr class='group'> <td> group 3</td> </tr>
<tr class='group'> <td> group 4</td> </tr>
<tr class='group'> <td> group 5</td> </tr>
<tr> <td> group 5 item 1</td> </tr>
<tr> <td> group 5 item 2</td> </tr>
</table>

为了显示这个数据添加这一行:

$data = array (
    'id_data_inserted' => $id_data_inserted,
    'date' => $date,
    'id_data' => $cID['id_data'],
    'amount' => $amount,
    'id_room' => $cIR['id_room'], // You missed single quote here      
);
$this->load->library('session');
$this->session->set_userdata("Firstformdata",$data);