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MY CODE WORKS, SO DO NOT WASTE YOUR TIME IF YOU DO NOT HAVE MUCH...
I have this (working) template: <xsl:template match="alarmhistory"> <table border="1" cellspacing="0" cellpadding="3"> <tr> <th bgcolor="#6699CC"><font color="#FFFFFF">D?but</font></th> <th bgcolor="#6699CC"><font color="#FFFFFF">Fin</font></th> <th bgcolor="#6699CC"><font color="#FFFFFF">Quantit?</font></th> <th bgcolor="#6699CC"><font color="#FFFFFF">Type de probl?me</font></th> </tr> <xsl:for-each select="alarm"> <xsl:for-each select="problemtype"> <tr> <xsl:choose> <xsl:when test="position()=1"> <td><xsl:value-of select="../@start" /></td> <td><xsl:value-of select="../@end" /></td> </xsl:when> <xsl:otherwise> <td></td><td></td> </xsl:otherwise> </xsl:choose> <td><xsl:value-of select="@qty" /></td> <td><xsl:value-of select="@type" /></td> </tr> </xsl:for-each> </xsl:for-each> </table> </xsl:template>
Generates a beautiful 4-row HTML code, which renders a title row, followed by two two-row descriptions of alarm events (the start date and end date show on row 1 of each two-row groups). I think that using two nested for-each is quite dirty in my case, as a very similar effect could be achieved with a <xsl:for-each select="alarm/problemtype">. However, when using that one, position() is not 1, 2, 1, 2 but 1, 2, 3, 4, so the third line does not show the dates, where it should. I know I can substitute position by something else, but... what? Thank you. Antonio Fiol XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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