How to call an xslt transform into an xsl stylesheet
I have written an xslt that reads some xml file names and does some operations on them.
I use a for-each to work them one-by-one. I have each path inside a parameter $path.But now I would like to output the result of applying an external stylesheet to those files. I would write something like
<div> <something like xsl-transform($extern-xslt,$path)> </div>
to have the result tree of the transformation inside t开发者_运维技巧he main html output. It is possible?
You could import the external stylesheet to your main stylesheet with xsl:import, and then just apply templates to that external XML file, which you can load with the document function.
<div><xsl:apply-templates select="document($path)"/></div>
If the templates in the external stylesheet would collide with the templates in the main stylesheet, you can use a different mode for them.
You can use the document() XPath function to load an external XML file. This returns a node-set which can be parsed with a <xml:apply-templates> call. Including an external stylesheet can be accomplished by using an <xsl:include> tag.
<xsl:include href="$external"/>
<xsl:apply-templates select="document($path)"/>
See also the documentation for document()
I have written an xslt that reads some xml file names and does some operations on them. I use a for-each to work them one-by-one. I have each path inside a parameter $path.
But now I would like to output the result of applying an external stylesheet to those files
The solution consists of these ingredients:
Use the standard XSLT
document()function to load and access the external XML document.Import the external stylesheet, using an
<xsl:import>instruction.The templates in the external stylesheet must be in a special mode, not used by the primary stylesheet.
At the place where the result of the "external transformation" is wanted, issue
<xsl:apply-templates>selecting the necessary nodes of the external document (usually the root node/, or the top element/*). The mode, specified on the<xsl:apply-templates>should be the same as the mode used in the external stylesheet.
Here is a small, simplified example (No external stylesheet is imported, the "external document" is embedded in the stylesheet, and no mode is used):
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns="http://www.w3.org/1999/xhtml"
xmlns:my="my:my"
>
<!-- <xsl:import href="myExternal.xsl"/> -->
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<my:div>
<h2>Weather</h2>
<p >It will be raining today</p>
</my:div>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="insertContents">
<xsl:apply-templates select="document('')/*/my:div/*"/>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on this XML document:
<html>
<h1>Today's News </h1>
<insertContents/>
</html>
the desired result is produced:
<html>
<h1>Today's News </h1>
<h2 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:my="my:my">Weather</h2>
<p xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:my="my:my">It will be raining today</p>
</html>
Note that the extraneous namespaces above are only due to the simplifications of this example -- they will not be generated if the external XML document were in its own file.
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