My requirement is as follows.

I have to merge 10 XML Files.

the input xml file structure is

<?xml version="1.0"?>
<document>
   <title>tile</title>
   <location>Geneva</location>
   <date>April 2003</date>
</document>

the output would be
<?xml version="1.0"?>
<rootnode>
<document>
   <title>tile</title>
   <location>Geneva</location>
   <date>April 2003</date>
</document> //from ten files
</rootnode>


i have written .xsl file which combines all the XML files.but i don't know how to create the <rootnode> rootelement in the .xsl.

please help me.
thanks in advance
Rajraman.N

Show us your code, and we can show you where to add the element constructor.

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

The following is my .xsl code

?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
<xsl:output method="xml" omit-xml-declaration="no" encoding="UTF-8"/>

<xsl:template match="/">
<xsl:for-each select="index/file">
       <xsl:apply-templates select="document(@filename)/root/document"/>
     </xsl:for-each>
</xsl:template>

<xsl:template match="root/document">
<xsl:copy>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>

<xsl:template match="title">
<xsl:copy>
   <xsl:apply-templates/>
   </xsl:copy>
</xsl:template>

<xsl:template match="location">
<xsl:copy>
  <xsl:apply-templates/>
  </xsl:copy>
</xsl:template>

<xsl:template match="date">
<xsl:copy>
   <xsl:apply-templates/>
   </xsl:copy>
</xsl:template>

</xsl:stylesheet>

Thanks in advance
N.Rajaraman

--

Joe (Microsoft MVP - XML)

Thanks Joe. Its working fine.

I have one more help.

If i stored all XML files in one folder, the following xsl is working fine.

.XSL code
<xsl:template match="/">
<xsl:for-each select="index/file">
       <xsl:apply-templates select="document(@filename)/root/document"/>
     </xsl:for-each>
</xsl:template>

.XML Code
<?xml version="1.0"?>
<index>
   <title>XML Files</title>
   <file filename="geneva.xml"></file>
   <file filename="london.xml"></file>
  </index>

if the each files are stored in sub folders under one root folder, what changes i have to make in .xsl and .xml files.

Thanks in advance.
Rajraman.N

When the document() function sees a relative URI, it resolves it against some base URI. If the argument to the document() function is a node, as in your example, then it uses the base URI of that node. In practice (given a relative URI such as geneva.xml) this means it looks in the directory containing the XML document that contains the relative URI. If the argument to the document() function is a string, on the other hand, then it is resolved relative to the stylesheet.

In XSLT 1.0 there are two ways you can change this behaviour. You can specify a base URI in the second argument of the document() function (see the spec for details) or you can modify the relative URI by string manipulation, for example concat("../subdir/", @filename).

In 2.0 there are many more choices, because you have access to functions such as base-uri() and resolve-uri() than give you a lot more control.

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