Thats the xml data, its a work in progress, so dont be too harsh on it
:) Its generated and transformed in PHP5 to output HTML.
releases is the data for the loop, releasepage is whats being matched
by my xsl:template element at the moment.
<?xml version="1.0"?>
<page>
<header stylesheet="header.xsl"/>
<body stylesheet="releases.xsl">
<releasepage>
<releases>
<row iteration="0">
<date>2005-01-01</date>
<name>Release Name 1</name>
</row>
<row iteration="1">
<date>2005-01-02</date>
<name>Release Name 2</name>
</row>
<row iteration="2">
<date>2005-01-03</date>
<name>Release Name 3</name>
</row>
<row iteration="3">
<date>2005-01-04</date>
<name>Release Name 4</name>
</row>
<row iteration="4">
<date>2005-01-05</date>
<name>Release Name 5</name>
</row>
<row iteration="5">
<date>2005-01-06</date>
<name>Release Name 6</name>
</row>
</releases>
</releasepage>
</body>
<footer stylesheet="footer.xsl"/>
</page>
Thanks for the comment btw Jay, i didnt notice another reply untill i
had sent the mail before :)
On Fri, 18 Mar 2005 10:29:24 -0500, Maria Amuchastegui
<mamuchastegui(_at_)to(_dot_)epost(_dot_)ca> wrote:
Can you post the XML data?
Maria
-----Original Message-----
From: Chris [mailto:phatfish(_at_)gmail(_dot_)com]
Sent: Friday, March 18, 2005 10:19 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Can i use apply-templates to match a xsl:template eleme
nt?
Hi thanks, i did look at call-template before in my testing but discarded it
because it didnt output my foreach loop in the template that i was calling.
But it does seem thats what i should be using, although im not sure why the
loop isnt being outputted -- and reading the specifications didnt really
make it any clearer :)
Bellow was the "main" template that i was hoping to output, it just contains
a test for-each loop.
Is it possible to have the template im calling output its content as i
wanted?
<?xml version="1.0" encoding="UTF-8"?>
<xsl:transform version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html"/>
<xsl:include href="layout.xsl"/>
<xsl:include href="header.xsl"/>
<xsl:template match="releasepage" name="main" >
<div>A table for the main template</div>
<table>
<tbody>
<xsl:for-each select="releases/row">
<tr>
<xsl:for-each select="date">
<td>
<xsl:apply-templates/>
</td>
</xsl:for-each>
<xsl:for-each select="name">
<td>
<xsl:apply-templates/>
</td>
</xsl:for-each>
</tr>
</xsl:for-each>
</tbody>
</table>
</xsl:template>
</xsl:transform>
On Fri, 18 Mar 2005 08:45:45 -0500, Maria Amuchastegui
<mamuchastegui(_at_)to(_dot_)epost(_dot_)ca> wrote:
You could do that with a named template:
<?xml version="1.0" encoding="UTF-8"?> <xsl:transform version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html"/>
<xsl:template match="page">
<html>
<head>
</head>
<body>
<div id="header">
<xsl:call-template name="header"/>
</div>
<div id="main">
<xsl:call-template name="main"/>
</div>
</body>
</html>
</xsl:template>
<xsl:template name="header">
<!-- do stuff here -->
</xsl:template>
<xsl:template name="main">
<!-- do stuff here -->
</xsl:template>
</xsl:transform>
-----Original Message-----
From: Chris [mailto:phatfish(_at_)gmail(_dot_)com]
Sent: Friday, March 18, 2005 8:39 AM
To: XSL List
Subject: [xsl] Can i use apply-templates to match a xsl:template element?
I would like the output of a xsl:template element to appear in a
specific location in stylesheet. Can i use apply-templates to match
the template i want and have it output there?
eg:
<?xml version="1.0" encoding="UTF-8"?> <xsl:transform version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html"/>
<xsl:template match="page">
<html>
<head>
</head>
<body>
<div id="header"><xsl:apply-templates
select="header"/></div>
<div id="main"><xsl:apply-templates
select="main"/></div>
</body>
</html>
</xsl:template>
</xsl:transform>
This file will be included into my stylesheets and used as a base
layout. I would like the xsl:template with name="main" to output into
the location above. But im not sure if this is allowed, does select
only relate an element in the source xml document?
I hope you can see what im trying to do, Thanks :)
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