Re: FW: How to navigate the tree from a selected node to the root?

Hello Nirmala,

I need a little enhancement.

<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="try.xsl"?>

<books>
<book name="name2">
        <otherdetails price="10"/>
        <otherdetails price="25"/> <!-- Newly added -->
</book>

<book name="name1">
        <otherdetails price="20"/>
        <otherdetails price="35"/> <!-- Newly added -->
</book>
</books>

Now applying the stylesheet as you have mentioned, I get the output as

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="try.xsl"?>
<books>
  <book name="name2">
        <otherdetails price="10"/>
        <otherdetails price="25"/>
</book>
</books>

My requirement is that I want to get

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="try.xsl"?>
<books>
  <book name="name2">
        <otherdetails price="10"/>
</book>
</books>

The sibling of otherdetails also should not be selected.

Hence I tried the stylesheet like

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
        <xsl:output method="xml" indent="yes"/>

<xsl:template match="books">
    <xsl:copy>
      <xsl:apply-templates select="@* | book[otherdetails/@price = '10']" />
    </xsl:copy>
  </xsl:template>

  <xsl:template match="@*|node()">
        <xsl:if test="node()/otherdetails/@price='10'">   <!-- I know, this is 
not
correct, but i need something like this -->


remove the <xsl:if/>

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


and add a further template matching on book:

<xsl:template match="book">
  <xsl:copy>
    <xsl:apply-templates select="@* | otherdetails[(_at_)price = '10']"/>
  </xsl:copy>
</xsl:template>

</xsl:stylesheet>

Basically, I should be able to differentiate within the template
match="@*|node(), whether
it is an attribute or a book node. If it is a book node and if
otherdetails/@price is not 10,
then that should not be included in the output xml.

Can you please tell me how could I do that?

Thanks in advance,
Best Regards,
Nirmala


So it's only a further special case in the identity transformation.

Hope this helps,

Joerg


XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list