RE: Multiple Rows in a Table / Same Element

Thanks Mike Brown and Mike Kay for your help.  I was able to get the rows I
needed but is there a way to get each individual date within it's own <TD>
tags??  Currently the days are all in one <TD> tag for each row.  Thanks
Again!
Rebecca Pinto 


Mike Brown wrote:

Be careful about the use of mod and position(); the first node is at
position 1, and you want positions 1, 8, 15, etc., not 0, 7, 14 -- so it
would be position() mod 7 = 1.

<xsl:template match="month">
  <xsl:for-each select="day[position() mod 7 = 1]">
    <tr>
      <xsl:apply-templates select=". | following-sibling::day[position()
&lt; 7]"/>
    </tr>
  </xsl;for-each>
</xsl:template>

This is what I used:

<xsl:template match="Month">
<xsl:for-each select="day[position() mod 7 = 1]">
<TR>
	<TD><xsl:apply-templates select="./@date |
following-sibling::day[position() &lt; 7]/@date" /></TD>
</TR>
</xsl:for-each>
</xsl:template>

And the output is this:

<TABLE border="1">
<TR>
<TD>1234567</TD>
</TR>
<TR>
<TD>891011121314</TD>
</TR>
<TR>
<TD>15161718192021</TD>
</TR>
<TR>
<TD>22232425262728</TD>
</TR>
<TR>
<TD>2930123</TD>
</TR>
</TABLE>


Original XML:
<Month>            
>             <day date="1" />
>             <day date="2"/>
>             <day date="3" />
> [...]
>             <day date="31"/>
> </Month>  


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list