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Subject:
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XSL Sorting using variable
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Posted By:
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mannabaron
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Post Date:
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2/14/2006 1:45:23 PM
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Hello, I have to post my question again because the first time I forgot (or I was to stupid) to explain someting.
Thank you very much to Joe and Michael Kay for answering my first question.
Assuming the following xml data:
<shop_articles>
<article id="1" order="50" show="yes">
<article_groups>
<article_group>
<name>Group1</name>
<value>value of group1</value>
</article_group>
<article_group>
<name>Quality</name>
<value>good</value>
</article_group>
...
</article_groups>
<article_price>400</article_price>
</article>
<article id...
</article>
</shop_articles>
I am using a <xsl:for-each > instruction to loop through all articles. When I want to sort the articles by the price it works fine with <xsl:sort select="*[name()=$article_sorting]" /> after the <xsl:for-each> instruction when the variable $article_sorting contains the value 'article_price'.
But I donīt know how to set the variable $article_sorting when I want to sort the articles by the <value> of <article_group>īs with the <name>Quality</name>.
Any one have an idea?
Thanks you very much in advance.
Greetings, Wolfgang
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Reply By:
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mhkay
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Reply Date:
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2/14/2006 2:18:10 PM
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<xsl:sort select="*[name()=$article_sorting]" /> allow you to sort by the value any child element, but it doesn't allow you to supply a completely arbitrary sort key at run-time. If you want a completely general solution, consider <xsl:sort select="saxon:evaluate($x)"/> where $x can hold any XPath expression - the trouble is this isn't available in all products.
If you don't want a completely general solution but just one of a number of possible sort keys, you can try multiple sort keys:
<xsl:for-each...
<xsl:sort select="*[name()=$article_sorting]" />
<xsl:sort select="article_groups[$article_sorting='']/article_group[name="Quality"]/value"/>
If $article-sorting is "" the first sort key has no effect, if it is not "" then the second sort key has no effect.
Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference
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Reply By:
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mannabaron
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Reply Date:
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2/15/2006 5:09:57 AM
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Hi Michael, thank you very much for your help, that was the solution for me. I know now a little bit more of XSL :)
Greetings Wolfgang
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Reply By:
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mannabaron
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Reply Date:
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2/15/2006 5:11:43 AM
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That was what I was searching for.
Thank you very much Michael.
Greetings, Wolfgang
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Reply By:
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mannabaron
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Reply Date:
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2/15/2006 5:13:20 AM
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Hi Michael, thank you very much again.
That was what I was looking for.
Greetings, Wolfgang
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