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Subject: [xsl] sorted table elements From: Saverio Perugini <sperugin@xxxxxxxxxxxxxxxx> Date: Thu, 3 Oct 2002 10:10:29 -0400 (EDT) |
Hello,
I'd like to produce a 4 column HTML table where each <td>
element is alphabetically sorted from cell 1,1 to cell n,4
(where n is the required number of rows).
The following is an attempt:
<table>
<xsl:for-each select="*[position() mod 4 = 1]">
<xsl:sort data-type="text" select="name()"/>
<tr>
<xsl:for-each select=".|following-sibling::node()[position() < 4]">
<td>...</td>
</xsl:for-each>
</tr>
</xsl:for-each>
</table>
This code only sorts all the *[position() mod 4 = 1] elements.
Therefore, only column one is sorted, whereas I wanted the entire
table (an linear list from cell 1,1 to cell n,4 sorted). Can this be done?
Thank You and Best Regards,
S. Perugini
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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