Subject: [xsl] How can I see the resulting XML structure after a transformation from one XML structure to another. From: Rashmi Rubdi <dev_subscriptions@xxxxxxxxx> Date: Sun, 12 Nov 2006 09:42:14 -0800 (PST) |
I'm trying to perform a 2 step transformation in my JSP file. I have a XML structure for a document where all links are under <document_links>, the XML also contains <paragraph> nodes. I want to transform this XML document so that all nodes except <paragraph> nodes are in the transformed XML. I want to test if transforming from one XML structure to another was done correctly before using the output of the transformation. But there doesn't seem to be a way in JSLT to see the xml node structure. I tried printing it with <x:out but it only outputs the text values and not the xml nodes themselves. I also tried to see the output of the transformation by performing clien-side transformation with IE6.0, but this transformation also shows only the text nodes. How can I see the *transformed XML structure* ? For example: I want to transform the following XML structure document.xml <?xml version="1.0" encoding="UTF-8"?> <document> <!-- Commented out , used for IE6.0 transformation. <?xml-stylesheet type="text/xsl" href="get_links.xsl"?> --> <document_links> <link> <id>1234</id> <url><![CDATA[http://www.test.com]]></url> </link> </document_links> <paragraph> <![CDATA[ test text ]]> </paragraph> </document> to this structure: (just trying to extract the XML with the link nodes, excluding paragraph nodes) <?xml version="1.0" encoding="UTF-8"?> <document> <!-- Commented out , used for IE6.0 transformation. <?xml-stylesheet type="text/xsl" href="get_links.xsl"?> --> <document_links> <link> <id>1234</id> <url><![CDATA[http://www.test.com]]></url> </link> </document_links> </document> This is the XSL: get_links.xsl <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" omit-xml-declaration="no" indent="no"/> <xsl:template match="/"> <xsl:apply-templates select="document"/> </xsl:template> <xsl:template match="document"> <document> <xsl:apply-templates select="document_links"/> </document> </xsl:template> <xsl:template match="document_links"> <document-links> <xsl:copy-of select="."/> </document-links> </xsl:template> </xsl:stylesheet> Snippet from JSP: <c:import url="/document.xml" var="xml"/> <c:import url="/get_links.xsl" var="xsl"/> <x:transform xml="${xml}" xslt="${xsl}" var="links_xml"/> <x:out select="$links_xml"/> The above <x:out only shows 1234 http://www.test.com and not the XML structure. I'm not sure if this is the right way to test the output of XML transformation Any help is appreciated. Regards Rashmi
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