parse multiple .XSD files with XSLT

chaostimmy · August 14th, 2009, 09:35 AM

Hi!
I have a problem with solving this problem,
i have 5 XSD Files (SCHEMA) where i want to get all <documentation> elements from and write them in one new XML File.
How can i do that ?
My problem is that i dont have any XML file which initialises the XSLT File...

Martin Honnen · August 14th, 2009, 10:08 AM

With XSLT 2.0 you can run a named template as the initial template:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:xs="http://www.w3.org/2001/XMLSchema"
  version="2.0">

  <xsl:template name="main">
     <root>
       <xsl:copy-of select="(document('schema1.xsd'), document('schema2.xsd'), document('schema3.xsd'), document('schema4.xsd'), document('schema5.xsd'))//xs:documentation"/>
     </root>
   </xsl:template>

</xsl:stylesheet>

You can then run your stylesheet with your XSLT processor and designate the template with name 'main' as the initial template, for instance with option -it:main for Saxon and its command line usage.

chaostimmy · August 15th, 2009, 06:48 AM

okay so far ive already been,
i did it with <xslt:variable name="file" select="doc('schema.xsd')">

i dont understand how to run that XSLT without an real input XML ...
In my Browser he needs an XML for it,
and Eclipse Oxygen parser wants an XML to...

Martin Honnen · August 15th, 2009, 07:28 AM

As I said, with XSLT 2.0 you do not need a primary input document, you can instead define a named template and start processing with that named template.
With Saxon there is a command line option -it:templatename you can use, with AltovaXML tools you can use a command line option /n templatename so for the stylesheet I posted you would use e.g.
altova.exe /xslt2 sheet.xsl /n main
with Saxon something like
java -jar dir/saxon9.jar -xsl:sheet.xsl -it:main

I am not familiar with Eclipse or Oxygen, ask in a forum dedicated to that software if you don't find a setting yourself.

If you use XSLT 1.0 then you do not have that option, in that case you either need to provide a dummy XML input or you can choose one of the schema documents as the primary input document.

chaostimmy · August 15th, 2009, 07:36 AM

ah okay now i know what youre talking about! BiG THANXXX

chaostimmy · August 17th, 2009, 08:24 AM

i tried it with <xsl:copy-of select="doc('filename.xsd')//documentation"/>
but it won't work ...

<xsl:copy-of select="doc('filename.xsd')//text()"/> does work....

whhhyyy ??

Martin Honnen · August 17th, 2009, 08:28 AM

Well the elements are in the namespace http://www.w3.org/2001/XMLSchema so bind a prefix to that namespace and use the prefix, as I did in my sample:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:xs="http://www.w3.org/2001/XMLSchema"
  version="2.0">

  <xsl:template name="main">
     <root>
       <xsl:copy-of select="(document('schema1.xsd'), document('schema2.xsd'), document('schema3.xsd'), document('schema4.xsd'), document('schema5.xsd'))//xs:documentation"/>
     </root>
   </xsl:template>

</xsl:stylesheet>

mhkay · August 17th, 2009, 08:32 AM

Martin wrote ...//xs:documentation, so why did you expect ...//documentation to work? The element is in a namespace, so like all elements in a namespace, the name must be prefixed in an XPath expression.

chaostimmy · August 17th, 2009, 09:58 AM

Okay, seems like i didnt understand that basic thing.. i hopefully do now!

first i had this Namespace:

Code:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns="http://www.w3.org/2001/XMLSchema"
    version="2.0">

so i thought //documentation shall work ..

now i use

Code:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    xmlns="http://www.w3.org/2001/XMLSchema"
    version="2.0">

so i dont have to mark all xml elements with "xs:", right?
any other suggestions for it ?? Or is there an error in reasoning again ?

here is now my working "make documentation xml" xslt code:

Code:

<xsl:template match="filelist">
                <documentations>
                    <xsl:for-each select="file">
                        <xsl:if test="doc-available(text())">
                            <xsl:variable name="data" select="doc(text())"/>
                            <xsl:for-each select="$data//node()/@name">
                                    <documentation>
                                        <name><xsl:value-of select="."/></name>
                                        <text><xsl:value-of select="parent::node()[//xs:documentation]"/></text>
                                    </documentation>
                            </xsl:for-each>
                        </xsl:if>
                    </xsl:for-each>
                </documentations>
</xsl:template>

with the input XML:

Code:

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="create_doc_XML.xsl" ?>
<filelist>
    <file>sourceData.xsd</file>
    <file>test-cases.xsd</file>
    <file>uoms.xsd</file>
</filelist>

i thought taking this xml with filenames would be a good way to make this program more dynamic..

now JUDGEMENT plz

by the way you guys are GREAT! big THANX

mhkay · August 17th, 2009, 10:02 AM

No, the default namespace in the stylesheet does not apply to unprefixed names in XPath expressions.

You can set a default namespaces for such names in XSLT 2.0 using the default-xpath-namespace attribute.