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RE: How to sort this xml out
- To: xsl-list at lists dot mulberrytech dot com
- Subject: RE: [xsl] How to sort this xml out
- From: Dan Diebolt <dandiebolt at yahoo dot com>
- Date: Wed, 9 May 2001 07:57:46 -0700 (PDT)
- Reply-To: xsl-list at lists dot mulberrytech dot com
The enclosed XML produces the following output:
Title
- first ReplyTitle
- Reply SecondTitle
Second Title
You may have to play with the <content> tag placement as I wasn't
clear if you wanted one outer content or one for each message
thread.
Regards,
Dan
-----------------
File: 9May2001Messages.xml
<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="9May2001Messages.xsl"?>
<Main>
<Message>
<id>15</id>
<parentID>0</parentID>
<title>Title</title>
</Message>
<Message>
<id>17</id>
<parentID>0</parentID>
<title>Second Title</title>
</Message>
<Reply>
<id>16</id>
<parentID>15</parentID>
<title>first ReplyTitle</title>
</Reply>
<Reply>
<id>18</id>
<parentID>15</parentID>
<title>Reply SecondTitle</title>
</Reply>
</Main>
File: 9May2001Messages.xsl
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<xsl:apply-templates select="Main"/>
</xsl:template>
<xsl:template match="Main">
<content>
<xsl:for-each select="Message[parentID='0']">
<xsl:value-of select="title"/><br/>
<xsl:variable name="ID" select="id"/>
<xsl:for-each select="../Reply[parentID=$ID]">
- <xsl:value-of select="title"/><br/>
</xsl:for-each>
</xsl:for-each>
</content>
</xsl:template>
</xsl:stylesheet>
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