How to get an absolute path to the directory of the XSL file?

Question

My Schema.xsd file is located in the same directory with the .xsl file. In the .xsl file I would like to generate a link to Schema.xsl in the generated output. The generated output is located in different directories. Currently I do it like this:

   <xsl:template match="/">
     <root version="1.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
           xsi:noNamespaceSchemaLocation="../../../Schema.xsd">
     <!-- . . . -->

However this forces the generated output to be located 3 levels under the directory of Schema.xsd. I would like to generate an absolute path to the schema in the output, so the output could be located anywhere.

Update. I use XSLT 1.0 (XslCompiledTransform implementation in .NET Framework 4.5).

Why don't you pass the directory of your XSL file as a parameter to the XSLT process? — Marcus Rickert, Sep 15 '14 at 10:14
@MarcusRickert, thank you for the hint. I am new to XSLT. I will have to research if it is possible to do in C# using `XslCompiledTransform`, because I run the xsl file this way. — Andrej Adamenko, Sep 15 '14 at 10:55

score 3 · Answer 1 · answered Sep 16 '14 at 00:12

XSLT 2.0 Solution

Use the XPath 2.0 function, resolve-uri():

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes"
              omit-xml-declaration="yes"
              encoding="UTF-8"/>
  <xsl:template match="/">
    <root version="1.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
          xsi:noNamespaceSchemaLocation="{concat(resolve-uri('.'), 'Schema.xsd')}">
    </root>
  </xsl:template>

</xsl:stylesheet>

Yields, without parameter passing and regardless of the input XML:

<root xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
      version="1.0"
      xsi:noNamespaceSchemaLocation="file:/c:/path/to/XSLT/file/Schema.xsd"/>

+1. I use XSLT 1.0, however this answer could be useful for others who has simular problem in 2.0. Or may be I can switch to 2.0 later. — Andrej Adamenko, Sep 16 '14 at 05:35

score 1 · Answer 2 · edited May 23 '17 at 12:21

This is a sketch of how to do it (also see Passing parameters to XSLT Stylesheet via .NET).

In your C# code you need to define and use a parameter list:

XsltArgumentList argsList = new XsltArgumentList();
argsList.AddParam("SchemaLocation","","<SOME_PATH_TO_XSD_FILE>");

XslCompiledTransform transform = new XslCompiledTransform();
transform.Load("<SOME_PATH_TO_XSLT_FILE>");

using (StreamWriter sw = new StreamWriter("<SOME_PATH_TO_OUTPUT_XML>"))
{
    transform.Transform("<SOME_PATH_TO_INPUT_XML>", argsList, sw);
}

Your XSLT could be enhanced like this:

...
<xsl:param name="SchemaLocation"/> <!-- this more or less at the top of your XSLT! -->
...

<xsl:template match="/">
   <root version="1.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:noNamespaceSchemaLocation="{$SchemaLocation}">
   ...
   ...
</xsl:template>
....

How to get an absolute path to the directory of the XSL file?

2 Answers2

XSLT 2.0 Solution