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Topic

eg bert

Subject: Newbie prob - Elements with same name as parent...
Author: eg bert
Date: 26 Sep 2006 10:37 AM

Hi

I'm reading through a 'todo' list generated by an app. called ToDoList

The 'basic' format of the file is: -

<?xml version="1.0" encoding="UTF-8"?>
<todolist>
<task id="1" title="Task 1">
<task id="1a" title="Task 1a"/>
<task id="1b" title="Task 1b"/>
</task>
<task id="2" title="Task 2">
<task id="2a" title="Task 2a"/>
</task>
</todolist>

So each <task> can have multiple sub tasks

I want to read through the file and output all tasks along with their attributes

I have got this far..

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:fn="http://www.w3.org/2004/07/xpath-functions" xmlns:xdt="http://www.w3.org/2004/07/xpath-datatypes">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/todolist">
<xsl:for-each select="task">
<xsl:sort order="ascending" select="@title"/>
<BR />Top:
<xsl:value-of select="@title"/>
(<xsl:value-of select="@id"/>)
<BR />
<xsl:value-of select="@comments"/>
<xsl:apply-templates select="task"/>
</xsl:for-each>
</xsl:template>
<xsl:template match="task">
<BR />Sub:
<xsl:value-of select="@title"/>
(<xsl:value-of select="@id"/>)
<BR />
<xsl:value-of select="@comments"/>
</xsl:template>
</xsl:stylesheet>

Output: -

Top: Task 1 (1)

Sub: Task 1a (1a)

Sub: Task 1b (1b)

Top: Task 2 (2)

Sub: Task 2a (2a)

but there must be a better way of doing this without repeating myself

Any help is greatly appreciated

Thanks, Eg.

James Durning

Subject: Newbie prob - Elements with same name as parent...
Author: James Durning
Date: 26 Sep 2006 10:48 AM

Add one line to this template:
<xsl:template match="task">
...
...
<xsl:apply-template match="task"/>
</xsl:template>

eg bert

Subject: Newbie prob - Elements with same name as parent...
Author: eg bert
Date: 26 Sep 2006 04:45 PM

Thanks very much James, that has picked up all the missing <tasks>.

Is it possible achieve this without outputting the attributes in the first template

<xsl:template match="/todolist">
<xsl:for-each select="task">
<xsl:sort order="ascending" select="@title"/>

+++ This block +++
<BR />Top:
<xsl:value-of select="@title"/>
(<xsl:value-of select="@id"/>)
<BR />
<xsl:value-of select="@comments"/>
--- This block ---
<xsl:apply-templates select="task"/>
</xsl:for-each>
</xsl:template>

so the 'task' templates does all the work?
Thanks Eg.

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