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Sushant Prabhu

Subject: retreiving ancestors
Author: Sushant Prabhu
Date: 16 Nov 2006 04:01 AM

Hi All

I have a scenario where i am retrieving all the ancestors for a particular node.

Here is my code
----
<xsl:for-each select="ancestor::*">
<xsl:value-of select="node_name"/>
<xsl:text>/</xsl:text>
</xsl:for-each>
----

Here though it retrieves all but not in the order I want. I mean the output is correct but fetch is in the opposite direction.
I want something like this
A
A-B
A-B-C
A-B-C-D

but what i am getting in return is the opposite
A-B-C-D
A-B-C
A-B
A

How can I get all the top parent listing first & then subsequent nested ones.

Do respond

Regards

James Durning

Subject: retreiving ancestors
Author: James Durning
Date: 16 Nov 2006 12:02 PM

would a sort work?
eg.
<xsl:sort select="0-position()"/>

Sushant Prabhu

Subject: retreiving ancestors
Author: Sushant Prabhu
Date: 16 Nov 2006 06:33 PM

Hi All

I tried with xsl:sort (ascending/descending) but did not help.
<xsl:for-each select="ancestor::*">
<xsl:sort select="." order="ascending" />

<xsl:value-of select="node_name"/>

</xsl:for-each>

I will give a glimpse of the xml which I have

*********************************

<node cid="1" oid="2">
<node_name>India</node_name>
<node cid="11" oid="4">
<node_name>Maharashtra</node_name>

<node cid="22" oid="8">
<node_name>Mumbai</node_name>

<node cid="33" oid="8">
<node_name>Dadar</node_name>
</node>
*********************************

Now what I want to print is something like this
India
India/Maharashtra
India/Maharashtra/Mumbai
India/Maharashtra/Mumbai/Dadar

But no matter what i tried its just giving me the output in the reverse order. Here is the code

<xsl:for-each select="ancestor::*">
<xsl:value-of select="node_name"/>
</xsl:for-each>
I applied the sort as suggested above but did not help.

Do let me know what I am missing here.

Regards

James Durning

Subject: retreiving ancestors
Author: James Durning
Date: 17 Nov 2006 09:41 AM
Originally Posted: 17 Nov 2006 09:37 AM

The outer loop surrounding that is important.
How do you get to that point?
Eg, this is one possible solution, but probably does not use the same method you use.

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format">
<xsl:output method="text"/>
<xsl:template match="/">
<xsl:for-each select="//node[not node]">
<xsl:for-each select="ancestor-or-self::*">
<xsl:for-each select="ancestor::*">
<xsl:value-of select="node_name"/>
<xsl:text>/</xsl:text>
</xsl:for-each>
<xsl:value-of select="node_name"/><xsl:text>
</xsl:text>
</xsl:for-each>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>

Sushant Prabhu

Subject: retreiving ancestors
Author: Sushant Prabhu
Date: 19 Nov 2006 07:53 PM

Hi James

I tried that solution but it did not help.

Find the code below which I am using

----------------------------------------------------
<xsl:template match="node">
<xsl:apply-templates select="node"/>
<xsl:for-each select="ancestor::*">
<xsl:value-of select="node_name"/>

<xsl:if test = "position() != '1'">
<xsl:text>/</xsl:text>
</xsl:if>
</xsl:for-each>
</xsl:template>
----------------------------------------------------

Do post your suggestions

Thanks & Regards

James Durning

Subject: retreiving ancestors
Author: James Durning
Date: 21 Nov 2006 10:34 AM

Move the apply-templates after the rest. You want the children's output after your own output.
<xsl:template match="node">
<xsl:for-each ....
</xsl:for-each>

<xsl:apply-templates select="node"/>
</xsl:template>

Sushant Prabhu

Subject: retreiving ancestors
Author: Sushant Prabhu
Date: 23 Nov 2006 07:30 PM

Hi James

Thanks a million

That worked gr8

Regards,

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