Compare values in different nodes

fann · August 26th, 2008, 09:11 AM

I need to copy the values of some nodes only if the values of these nodes != values in some other nodes.

example input:

<...>
<nodes>
<node>
AAAA
</node>
<node>
BBBB
</node>
...
</nodes>
<...>
<dont_copy>
<values>
<value>
AAAA
</value>
</values>
<...>

The output should be:

<node>
BBBB
</node>
<node>
...

It is somewhat difficult to explain but anyone know how to accomplish this?

mhkay · August 26th, 2008, 09:15 AM

Two template rules:

<xsl:template match="node">
<xsl:copy-of select="."/>
</xsl:template>

<xsl:template match="node[.=//dont_copy/values/value]"/>

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

Martin Honnen · August 26th, 2008, 09:16 AM

Code:

<xsl:copy-of
  select="//nodes/node[not(. = //dont_copy/values/value)]"/>

--
Martin Honnen
Microsoft MVP - XML

fann · August 27th, 2008, 02:42 AM

Hmm still cant get it to work, I think I was a bit unclear in my example input. So how about this one:

Code:

<root>
    <nodes>
        <node>
            <name>
                AAA
            </name>
            <value>
                1
            </value>
        </node>
        <node>
            <name>
                BBB
            </name>
            <value>
                2
            </value>
        </node>
        <node>
            <name>
                CCC
            </name>
            <value>
                3
            </value>
        </node>
        ...
    </nodes>
    ...
    <items>
        <item>
            <name>
                AAA
            </name>
        </item>
        <item>
            <name>
                CCC
            </name>
        </item>
        <item>
            ...
        </item>
        ...
    </items>
    ...
</root>

And the result should become something like:

Code:

<root>
    <nodes>
        <node>
            <name>
                BBB
            </name>
        </node>
        <node>
            ...
        </node>
        <node>
            ...
        </node>
    </nodes>
</root>

I think if I only have to compare to one node to not copy, I could get it to work but since it is a list of values that I have to compare to Id need some kind of loop right? I just started with XSLT yesterday (need it in my BizTalk solution) so if you guys could explain like you would to an idiot I would appreciate it!

fann · August 27th, 2008, 02:46 AM

Oops output should be something like:

Code:

<root>
    <nodes>
        <node>
            <name>
                BBB
            </name>
            <value>
                2
            </value>
        </node>
        <node>
            ...
        </node>
        <node>
            ...
        </node>
    </nodes>
</root>

Martin Honnen · August 27th, 2008, 06:30 AM

Here is a sample stylesheet:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="1.0">

  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@* | node()"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="root">
    <xsl:copy>
      <xsl:apply-templates select="@* | nodes"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="node[name = /root/items/item/name]"/>

</xsl:stylesheet>

When applied to this input XML

Code:

<root>
    <nodes>
        <node>
            <name>
                AAA
            </name>
            <value>
                1
            </value>
        </node>
        <node>
            <name>
                BBB
            </name>
            <value>
                2
            </value>
        </node>
        <node>
            <name>
                CCC
            </name>
            <value>
                3
            </value>
        </node>
    </nodes>
    <items>
        <item>
            <name>
                AAA
            </name>
        </item>
        <item>
            <name>
                CCC
            </name>
        </item>

    </items>

</root>

then the result with Saxon 6.5.5 is

Code:

<?xml version="1.0" encoding="utf-8"?><root><nodes>

        <node>
            <name>
                BBB
            </name>
            <value>
                2
            </value>
        </node>

    </nodes></root>

The stylesheet uses the identity transformation template and then has two additional templates, one for the 'root' element to make sure that only the 'nodes' elements are processed but not the 'items' elements, and one empty template for 'node[name = /root/items/item/name]' elements to ensure that these are not copied. That should suffice, no need for any loops.

--
Martin Honnen
Microsoft MVP - XML

fann · September 11th, 2008, 05:34 AM

Thanks, Ive been busy for a while so unfortunately I still haven't gotten this to work.

I use the following xslt (nodes where Addition_Name = addition_name shouldn't be copied):

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  targetNamespace="http://DataRegisterRequest"
  version="1.0">

    <xsl:template match="@* | node()">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="*[local-name()='Root']">
        <xsl:apply-templates select="//*[local-name()='ADDITIONS']"/>
    </xsl:template>

    <xsl:template match="*[local-name()='ADDITION'][*[local-name()='Addition_Name'] = //*[local-name()='KLNOrder']//*[local-name()='addition_name']]"/>

</xsl:stylesheet>

Now only the nodes that are supposed to be copied are actually copied. But there is still a problem with the namespaces. The resulting XML file has the same namespace as the original XML. I need the new XML to have a namespace defined by me. How can I accomplish this with my current xslt?

fann · September 11th, 2008, 05:36 AM

There is an error in the prev post. Here is my XSLT:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="1.0">

    <xsl:template match="@* | node()">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="*[local-name()='Root']">
        <xsl:apply-templates select="//*[local-name()='ADDITIONS']"/>
    </xsl:template>

    <xsl:template match="*[local-name()='ADDITION'][*[local-name()='Addition_Name'] = //*[local-name()='KLNOrder']//*[local-name()='addition_name']]"/>

</xsl:stylesheet>

samjudson · September 11th, 2008, 05:57 AM

For elements you'd do something like this:

Code:

<xsl:template match="node()">
        <xsl:element name="{local-name()}" namespace="http://newnamespace.com">
            <xsl:apply-templates select="@* | node()"/>
        </xsl:element>
</xsl:template>

and the same for attributes:

Code:

<xsl:template match="@*">
        <xsl:attribute name="{local-name()}" namespace="http://newnamespace.com">
            <xsl:value-of select="."/>
        </xsl:attribute>
    </xsl:template>

/- Sam Judson : Wrox Technical Editor -/

fann · September 11th, 2008, 06:25 AM

Here is how I managed to solve it:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:ns0="mynamespace"
  version="1.0">

    <xsl:template match="*[local-name()='Addition_Name']">
        <Addition_Name>
            <xsl:value-of select="."/>
        </Addition_Name>
    </xsl:template>

    <xsl:template match="*[local-name()='Addition_Value']">
        <Addition_Value>
            <xsl:value-of select="."/>
        </Addition_Value>
    </xsl:template>

    <xsl:template match="*[local-name()='Root']">
        <ns0:ADDITIONS>
            <xsl:apply-templates select="//*[local-name()='ADDITIONS']"/>
        </ns0:ADDITIONS>
    </xsl:template>

    <xsl:template match="*[local-name()='ADDITION'][*[local-name()='Addition_Name'] = //*[local-name()='KLNOrder']//*[local-name()='addition_name']]"/>

</xsl:stylesheet>

I find it really tricky to understand how XSLT works, but when I finally figure out how to do something in XSLT it just seems so obvious...