FW: [xsl] How to navigate the tree from a selected node to the root?

["Nirmala R" <nirmala.r@xxxxxxxxxx>]

Hi,

Can anyone tell me how can I get the following output?
Please refer the below mail.

Thanks in advance and regards,
Nirmala

-----Original Message-----
From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Nirmala R
Sent: Saturday, October 19, 2002 5:01 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE: [xsl] How to navigate the tree from a selected node to the
root?


Hi,

Thank you very much for the answer.
That answers my question. Thank you very much.

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



-----Original Message-----
From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of
Jarno.Elovirta@xxxxxxxxx
Sent: Friday, October 18, 2002 5:28 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE: [xsl] How to navigate the tree from a selected node to the
root?


Hi,

> <?xml version="1.0"?>
> <?xml-stylesheet type="text/xsl" href="try.xsl"?>
>
> <books>
> <book name="name2">
> 	<otherdetails price="10"/>
> </book>
>
> <book name="name1">
> 	<otherdetails price="20"/>
> </book>
> </books>
>
> and an xsl for the above as:
>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
> 	<xsl:output method="xml"/>
>
> 	<xsl:template match="/">
> 	<xsl:for-each select="books/book/otherdetails[@price='10']">
> 	<xsl:copy-of select="parent::*"/>
> 	</xsl:for-each>
>
> 	</xsl:template>
> </xsl:stylesheet>
>
> This gives the output as:
>
> <?xml version="1.0" encoding="UTF-8"?>
> <book name="name2">
>         <otherdetails price="10"/>
> </book>
>
> I want to get like this:
> <books>
> <book name="name2">
>         <otherdetails price="10"/>
> </book>
> </books>
>
> i.e. From this paricular selected node, i want to navigate
> till the root
> node and get the output.
> I will not know how many more ancestors are there to reach
> the root node.
>
> Can you please help me out for doing the same.

Will changing the approach to

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

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

be a suitable solution?

Cheers,

Jarno

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