问使用PHP问题将html表导出到可下载的excel

EN

<table>         
        <thead>
            <tr>
                <th colspan="5">Incoming</th>
            </tr>
            <tr>
                <th>From</th>
                <th>To</th>
                <th>Date</th>
                <th>Duration</th>
                <th>Status</th>
            </tr>
        </thead>
        <tbody>
            <tr>
                <td style="text-align: center;">942367233</td>
                <td style="text-align: center;">-</td>
                <td style="text-align: center;">15-06-2011 08:24</td>
                <td style="text-align: center;">00:00</td>
                <td style="text-align: center;">Abandoned</td>
            </tr>
            <tr>
                <td style="text-align: center;">935761500</td>
                <td style="text-align: center;">1956</td>
                <td style="text-align: center;">15-06-2011 09:20</td>
                <td style="text-align: center;">00:00</td>
                <td style="text-align: center;">Answered</td>
            </tr>
            <tr>
                <td style="text-align: center;">942367233</td>
                <td style="text-align: center;">1957</td>
                <td style="text-align: center;">15-06-2011 09:21</td>
                <td style="text-align: center;">02:16</td>
                <td style="text-align: center;">Answered</td>
            </tr>
        </tbody>
</table>

<form action="toexcel.php" method="post" target="_blank"
        onsubmit=\'$("#datatodisplay").val($("#data").html())\'>
        <input  type="image" src="/images/icons/Floppy-48x48.png" width="12" height="12">
        <input type="hidden" id="datatodisplay" name="datatodisplay" />
        </form>

header("Content-type: application/vnd.ms-excel; name='excel'");
header("Content-Disposition: filename=ficheroExcel.xls");  
header("Pragma: no-cache");  
header("Expires: 0");
echo $_POST['datatodisplay'];