用xpath如何得到一个node的position

[yoken]

<?xml version="1.0" encoding="GBK"?>
<table trClick="valid">
<tr>
<td style="display:none">XXCW0000000000</td>
<td width="205">abc</td>
<td width="75">2003-09-12</td>
<td style="display:none"/>
</tr>
<tr>
<td style="display:none">XXCW0000000004</td>
<td width="205" style="color:red">2</td>
<td width="75">2003-10-28</td>
<td style="display:none"/>
</tr>
<tr>
<td style="display:none">XXCW0000000005</td>
<td width="205" style="color:red">3</td>
<td width="75">2003-10-28</td>
<td style="display:none"/>
</tr>
<tr>
<td style="display:none">XXCW0000000006</td>
<td width="205" style="color:red">5</td>
<td width="75">2003-10-28</td>
<td style="display:none"/>
</tr>
<tr>
<td style="display:none">XXCW0000000007</td>
<td width="205" style="color:red">6</td>
<td width="75">2003-10-28</td>
<td style="display:none"/>
</tr>
<tr>
<td style="display:none">XXCW0000000008</td>
<td width="205" style="color:red">7</td>
<td width="75">2003-10-28</td>
<td style="display:none"/>
</tr>
<tr>
<td style="display:none">XXCW0000000014</td>
<td width="205" style="color:red">次</td>
<td width="75">2003-11-18</td>
<td style="display:none"/>
</tr>
<tr>
<td style="display:none">XXCW0000000013</td>
<td width="205" style="color:red">阿</td>
<td width="75">2003-11-18</td>
<td style="display:none"/>
</tr>
<tr>
<td style="display:none">XXCW0000000015</td>
<td width="205" style="color:red">波</td>
<td width="75">2003-11-18</td>
<td style="display:none"/>
</tr>
<tr>
<td style="display:none">XXCW0000000012</td>
<td width="205" style="color:red">111111</td>
<td width="75">2003-10-28</td>
<td style="display:none"/>
</tr>
</table>





通过/table/tr[1]/td[not(@style) or @style != 'display:none']可以得到两个td，但如何得到这两个td在parent tr中的position

[adjoin]

position()

[saucer]

try

count(preceding-sibling::td)

[yoken]

ok:
count(preceding-sibling::td) + 1

[yoken]

如
count(/table/tr[1]/td[not(@style) or @style != 'display:none'][1]/preceding-sibling::td) + 1