How to pass variables to collection(), document()

mphare · September 10th, 2008, 04:41 PM

I'm using saxon9 and ant 1.7 on a PC.

I have a XSLT that refers to a directory that has multiple XML files in it. The stylesheet runs against the information in these XML files.

I've done this using the collection() function, as:

Code:

<xsl:variable name="subtreeCmdsource" select="collection('/Development/structured-fm/source/?select=*.xml')"></xsl:variable>

This was ok for testing, but now I need to make the script flexible and allow for the directory to be passed in as a stylesheet parameter from ant. The parameter from ant part is easy, but I can't figure out how to use the parameter in the collection() function.

Code:

<xsl:variable name="subtreeCmdsource" select="collection('$targetDir/?select=*.xml')"></xsl:variable>

Doesn't work.

I know you're probably slapping your head going "what an idiot", but this is what made sense to me.

Is there a way to do this?

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Michael Hare

mhkay · September 10th, 2008, 05:13 PM

The argument to collection() is a string, and you can construct the string at run-time using an XPath expression in the same way as any other string, for example

collection(concat($dir, '?select=*.xml'))

Clearly if $targetDir appears inside a string literal as in your example then it's not going to be treated as a variable reference.

Michael Kay
http://www.saxonica.com/
Author, XSLT 2.0 and XPath 2.0 Programmer's Reference

mphare · September 11th, 2008, 10:15 AM

Thanks Michael!

That was just what I needed.

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Michael Hare

JessicaD · July 15th, 2009, 01:54 PM

I'm somewhat new to XSLT, so please forgive my ignorance.

My issue is similar, but the concat suggestion isn't working for me. I have an XML file with links to other XML files that contain snippets of content. I'm trying to insert those content snippets into the parent XML file where the link references occur. I created a variable that captures the link reference (if one exists) using this code:

Code:

<xsl:variable name="snippetpath">
  <xsl:choose>
    <xsl:when test="contains(item[@name='ContentSection']/value,'Snippet')" >	
      <xsl:value-of select="item[@name='ContentSection']/value/p/snippet/@dcrpath"/>	
    </xsl:when>
    <xsl:otherwise>
      <xsl:value-of select="item[@name='ContentSection']/value"/>	
    </xsl:otherwise>
  </xsl:choose>
</xsl:variable>

I know it's capturing the correct information because I can see the links using this code:

Code:

<xsl:value-of select="$snippetpath"/>

When I pass the $snippetpath variable to the document function using this code:

Code:

<xsl:value-of disable-output-escaping="yes" select="document('$snippetpath')//Properties/Data/Datum[@ID='D02']/DCR[@Type='ContentSnippet']/record/item[@name='ContentArea']/value/item[@name='ContentSubSection']/value/item[@name='ContentSection']/value"/>

I get a message that the resource cannot be found. But, if I hard code the path into the document function, it works:

Code:

<xsl:value-of disable-output-escaping="yes" select="document('../SNIP/SPSx10879')//Properties/Data/Datum[@ID='D02']/DCR[@Type='ContentSnippet']/record/item[@name='ContentArea']/value/item[@name='ContentSubSection']/value/item[@name='ContentSection']/value"/>

I can't figure out what I'm doing wrong. Any suggestions are greatly appreciated!

Martin Honnen · July 15th, 2009, 01:59 PM

You want document($snippetlink) without the quotes around the variable reference.

JessicaD · July 15th, 2009, 02:04 PM

THANK YOU!!! Of course it's the simplest of things... Thanks for your help!

mhkay · July 15th, 2009, 02:25 PM

Also, disable-output-escaping="yes" is best avoided unless you really know what you are doing. When used by a beginner, it generally indicates a serious misunderstanding of how to achieve something quite simple.

JessicaD · July 16th, 2009, 09:32 AM

Thanks for the advice. I inherited the code and the project, so I'm not sure what was done before me or why. But in this case where I'm trying to pull in XML from another single-sourced file, would DOE be appropriate since I want the nodes from the single-sourced content? How else can I get it to recognize and parse the tags?

Thanks in advance for your help!

Martin Honnen · July 16th, 2009, 09:44 AM

You would need to disable-output-escaping if you have text with escaped markup e.g. with

Code:

<foo><![CDATA[<p>...<p>...<br>...]]></foo>

we have a 'foo' element in the XML that has as its content escaped HTML markup. If we want the stylesheet to create HTML output with the contents of the 'foo' element we can use

Code:

<xsl:template match="foo">
  <xsl:value-of select="." disable-output-escaping="yes"/>
</xsl:template>

as long as the XSLT processor supports that for the output you use it with.

But if you have real XML nodes in your input as in

Code:

<foo><p>...</p><p>...<br /></p></foo>

then you don't need disable-output-escaping, you can simply copy nodes to the output:

Code:

<xsl:template match="foo">
  <xsl:copy-of select="node()"/>
</xsl:template>

JessicaD · July 16th, 2009, 10:09 AM

Thanks for the explanation. My input has real XML, so I removed the DOEs and changed the value-of to copy-of and it worked! Thanks so much!!