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Or, as was suggested to me on this list, you can use
<xsl::value-of select="//Archive[not(../Archive/@ID > @ID)]/@ID" /> Which I found a very compact and nice way of understanding the problem. Explanation: Look for the ID attribute of an Archive node for which it does NOT exist any ../Archive (sibling Archive) node whose ID is greater than its (the first node I referenced) ID. Did you understand my English? I do not... I'll rephrase: We look for the ID of the Archive node with the greatest ID. That is, we look for (the ID of) a node whose ID is greater or equal than any other. That is, we look for (the ID of) a node whose ID is NOT strictly lower than ANY of his siblings. That's better for my English skills, I believe ;-) Antonio Fiol Pfitzner, Jan wrote: Yes, you can. You can sort the Archive-nodes using the ID with xsl:sort and extract the first or last node of the sorted tree. <xsl:variable name="max"> <xsl:sort data-type="number" select="//Archive/@ID" order="descending"/> <xsl:value-of select="//Archive[1]/@ID"/> </xsl:variable> Attachment:
smime.p7s
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