Subject: Re: [xsl] sort by xsl:if and then sort using templates? From: "Max Bronsema" <max.bronsema@xxxxxxxxx> Date: Mon, 20 Nov 2006 15:05:37 -0800 |
I seem to not be able to wrap my head around how the XSLT processor works, I will be doing some reading and then getting back at this problem.
xsl:sort have to be direct children of xsl:for-each (or xsl:aply-templates) you can not nest in xsl:if (or anything else) Just put the test on your select predicate or outside the xsl:apply-templates
It's a bit hard to show exactly what you need to do as I think you over trimmed your stylesheet, but in general rather than
<xsl:apply-templates> <xsl:if test="something"> <xsl:sort ..../> </xsl:if> </xsl:apply-templates>
do either
<xsl:apply-templates select="*[something]> <xsl:sort.../> </xsl:sort>
or
<xsl:choose> <xsl:when test="something"> <xsl:apply-templates> <xsl:sort ..../> </xsl:apply-templates> </xsl:when> <xsl:otherwise> <xsl:apply-templates/> </xsl:otherwise>
depending on what you want to do if the predicate is false (no output, or no sorting, respectively)
David
Reply Reply to all Reply to allForward Forward Print Add Rashmi to Contacts list Delete this message Report phishing Show original Message text garbled? Rashmi Rubdi <dev_subscriptions@xxxxxxxxx> to xsl-list show details 12:53 pm (2 hours ago) Which version of XSLT/XPATH? I dont think you can compare dates with > in XPATH1.0
You can't nest xsl:sort inside xsl:if. xsl:if can only be nested under <xsl:apply-templates> and <xsl:for-each>
I think if you represent dates in the standard format for example: 2005-05-07T19:30:00 instead of 05/07/2005 you can simply sort dates without having to extract the substrings of year, month and day with reference to the code snippet you have provided.
like this: <xsl:apply-templates> <xsl:sort select="date"/> </xsl:apply-templates>
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