Question

I need to transform an XML file to another XML file, where the source file has a dynamic namespace set to xmlns="whatever". My XSLT runs fine without the namespace being in the file, but I get no output with the namespace. How can I cause the schema of the source file to be applied to the destination file?

All help is appreciated and thanks in advance!

EDIT:

I’m trying to copy the namespace uri over to the resulting file:

<xsl:param name="schema">
    <xsl:value-of select="namespace-uri()" />
</xsl:param>

<xsl:element name="root" namespace="$schema">

I have verified that schema is holding the correct value, but the problem is that the program appears to take this too literally:

<root xmlns="$schema">

Is this the right way to go about this?

EDIT x2:

I’ve implemented Alejandro’s suggestion of:

<xsl:element name="root" namespace="{$schema}"/> 

And that works for the most part, except for the fact that I have to put the namespace on every element or else I get the following structure in the result:

<root xmlns="NAMESPACE">
    <foo xmlns="">

etc.

Is there a way to blanket all of the elements with this namespace, other than putting namespace={$schema} on every single line? Bounty and accept for the best answer!

EDIT x3:
Better example:

If I do:

<xsl:element name="root" namespace="{namespace-uri()}>
  <xsl:element name="foo">
    <xsl:element name="bar">
    <!--etc-->
    </xsl:element>
  </xsl:element>
</xsl:element>

I get:

<root xmlns="NAMESPACE">
  <foo xmlns="">
    <bar>
    <!--etc-->
    </bar>
  </foo>
<root>

I would like to have them all under namespace NAMESPACE, so I did:

<xsl:element name="root" namespace="{namespace-uri()}>
  <xsl:element name="foo" namespace="{namespace-uri()}>
    <xsl:element name="bar" namespace="{namespace-uri()}>
    <!--etc-->
    </xsl:element>
  </xsl:element>
</xsl:element>

However this is ugly and tedious to type. Is there an easier way to blanket the namespace over all elements? (hopefully this clarifies what I need)

Answer 1

Suppose you have this XML input:

<root xmlns="survivors"> 
  <louis/> 
  <francis/> 
</root> 

Meaning that every element is under default namespace wich its URI is “survivors”.

As Welbog wrote you can select francis element with:

/*/*[local-name()='francis']

or

/*[local-name()='root']/*[local-name()='francis']

But, that also select francis element from these XML inputs:

<root xmlns="survivors" xmlns:n="no-survivors"> 
  <louis/> 
  <n:francis/> 
</root> 

or

<root xmlns="survivors"> 
  <louis/> 
  <francis xmlns="no-survivors"/> 
</root> 

You could also strengthen the predicate with some namespace URI. But, wich one? An option could be the default namespace for root element like:

/*/*[local-name()='francis'][namespace-uri()=namespace-uri(/*)]

Surely this make XPath expression very verbose.

In XSLT 2.0 you could use xsl:xpath-default-namespace attribute like:

<xsl:value-of select="/root/francis" xpath-default-namespace="survivors"/> 

But that’s not good for your case because you don’t know the URI in advance.

EDIT: xsl:element ‘s attributes are AVT (Attribute Value Template) so you need this:

<xsl:element name="root" namespace="{$schema}"/> 

Also, I recomend you to declare the param as a string data type (not RTF like now), something like:

<xsl:param name="schema" select="namespace-uri()"/> 

EDIT 2: Maybe I was not clear. You don’t need xsl:element/@namespace in every case. Following your statement that every element is in only one default namespace, this stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="@*|node()">
        <xsl:copy>
                <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="*[local-name()='bar']">
        <xsl:element name="newbar" namespace="{namespace-uri()}">
                <xsl:apply-templates select="@*|node()"/>
        </xsl:element>
    </xsl:template>
</xsl:stylesheet>

With this input:

<root xmlns="whatever">
    <foo/>
    <bar/>
</root>

Output:

<root xmlns="whatever">
    <foo></foo>
    <newbar></newbar>
</root>

Edit 2: I was showing you that when you are copying an element you are also copying the namespace needed for expand the QName. So, if want to transform this:

<root xmlns="whatever">
    <foo/>
    <bar/>
</root>

Into this:

<root xmlns="whatever">
    <foo>
        <bar/>
    </foo>
</root> 

You can use this stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="*">
        <xsl:copy>
            <xsl:apply-templates select="*[1]|following-sibling::*[1]"/>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>