Subject: RE: Multiple Rows in a Table / Same Element From: "Pinto, Rebecca" <rebecca.pinto@xxxxxxxxxxxxx> Date: Wed, 9 Aug 2000 15:35:41 -0400 |
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() < 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() < 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
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