How to get rid of root node and child node...

q1sevast · April 13th, 2012, 05:36 AM

Hi all,

I am not that skilled programming Stylesheets but I think that what I want to achieve is not that complicated. I have the following XML:

<?xml version="1.0" encoding="UTF-8"?>
<SIT:Wrapping xmlns:SIT="http://www.sit.com/xml/sitwrapping">
<SIT:Header>
<SIT:Operation>OperationName</SIT:Operation>
</SIT:Header>
<BOD:Body xmlns:BOD="http://www.sit.com/xml/body">
<BOD:Node>
<BOD:Element1>ElementValue</BOD:Element1>
<BOD:Element2>ElementValue</BOD:Element2>
</BOD:Node>
</BOD:Body>
</SIT:Wrapping>

Basically I need to get rid of the wrapping node together with the header so the desired XML would look like this:

<?xml version="1.0" encoding="UTF-8"?>
<BOD:Body xmlns:BOD="http://www.sit.com/xml/body">
<BOD:Node>
<BOD:Element1>ElementValue</BOD:Element1>
<BOD:Element2>ElementValue</BOD:Element2>
</BOD:Node>
</BOD:Body>

The only tricky part is that Body will not always be the same and its name will change depending on the type of the to-be-processed XML request. So i need the XSL to dinamically take care of that.

Hope you guys can help me.

Thanks all for your time,

Martin Honnen · April 13th, 2012, 05:41 AM

Well you need to explain to use how exactly the node to be copied is determined if the name and/or namespace can change. Do you want to copy the second child element of the root element for instance?

q1sevast · April 13th, 2012, 05:51 AM

Hi Martin,

Thanks for your answer. For the second root child, the namespace will always be the same but the name can be different. So basically yes, I want to keep always the second rott child element (and its child hearchy).

Regards,
Jose

Martin Honnen · April 13th, 2012, 05:56 AM

You could try

Code:

<xsl:template match="/">
  <xsl:copy-of select="*/*[2]"/>
</xsl:template>

then although I think you might complain then that the namespace declarations on the root are copied through.
If you don't want them then you need more code e.g.

Code:

<xsl:template match="/">
  <xsl:apply-templates select="*/*[2]"/>
</xsl:template>

<xsl:template match="*">
  <xsl:element name="{name()}" namespace="{namespace-uri()}">
    <xsl:copy-of select="@*"/>
    </xsl:apply-templates/>
  </xsl:element>
</xsl:template>

or if you use XSLT 2.0 you could make use of its options not to copy namespaces.

Martin Honnen · April 13th, 2012, 05:58 AM

To add a sample of XSLT 2.0

Code:

<xsl:template match="/">
  <xsl:copy-of select="*/*[2]" copy-namespaces="no"/>
</xsl:template>

q1sevast · April 13th, 2012, 08:19 AM

Thanks!!! Ill try out and let you know....

q1sevast · April 16th, 2012, 02:56 AM

Hi Martin,

Worked like a charm using XSL 2.0...Thanks a lot for your help.

Regards,
Jose