我想使用ajax更改表值

时间:2016-02-02 20:15:00

标签: javascript php jquery ajax codeigniter

我有一个选择选项,我选择一个,我想从数据库中更改tablo值。 在我的观点中:

<select id="Room" >
    <option value="Joseph">Joseph Room</option>
    <option value="Macro">Macro Lounge</option>
</select>
<table>
    <tr>
        <td>Day</td>
        <?php
        foreach ($posts as $post)
        {
            echo "<td>{$post->seancehour}</td>";
        }
        ?>
    </tr>
    <tr>
        <td><?php echo $post->seanceday; ?></td><td><?php echo $post->seancename; ?></td>
    </tr>
</table>

我使用ajax并想要更改dynamiccally表值。我的javascript是:

<script type="text/javascript">
    $(document).ready(function () {
        $('#Room').change(function () {
            var room = document.getElementById("Room").value;
            $.ajax({
                type: 'POST',
                data: {room: room},
                url: '<?php echo site_url('Homepage/index'); ?>',
                success: function (result) {
                    alert(result);//I dont know what has to be here.
                }
            });
        });
    });
</script>

在我的Controller中,我转到mymodel并获取数据库值。

public function index()
{
    $room = $this->input->post('room');

    if ($room == 'Macro')
    {
        $this->data['pasts'] = $this->SeanceModel->getMacro();
    }
    else
    {
        $this->data['posts'] = $this->SeanceModel->getJoseph();
    }
    $this->load->view('Homepage', $this->data);
}

例如,当我打开页面day = Sunday hour = 12 pm时,如果我更改了joseph或macro,则必须在表格中更改此值。

0 个答案:

没有答案