converting Table XML to HTML

kvbhaskar7 · November 4th, 2011, 07:11 PM

Hi,

I have the below XML table where i have to convert it to HTML. I need to consider the XML attributes to create my HTML table...

Can someone help me in XSLT creation..

<Table ID="Tab7" Float="Yes" Width="100" Style="Style1">
<Caption Language="En">
<CaptionNumber>Table 2.7</CaptionNumber>
<CaptionContent>
<SimplePara>My Table</SimplePara>
</CaptionContent>
</Caption>
<tgroup cols="2">
<colspec colnum="1" colname="c1" align="left" colwidth="37.64"/>
<colspec colnum="2" colname="c2" align="left" colwidth="62.36"/>
<thead>
<row>
<entry align="left" colname="c1"> <SimplePara>Employee Number</SimplePara>
</entry>
<entry align="left" colname="c2"> <SimplePara>Employee Name</SimplePara> </entry>
</row>
</thead>
<tbody>
<row> <entry align="left" colname="c1"> <SimplePara>1</SimplePara> </entry> <entry align="left" colname="c2"> <SimplePara>John</SimplePara> </entry> </row> <row> <entry align="left" colname="c1"> <SimplePara>2</SimplePara> </entry> <entry align="left" colname="c2"/> </row> <row> <entry align="left" colname="c1"> <SimplePara>3</SimplePara> </entry> <entry align="left" colname="c2"> <SimplePara>Tom</SimplePara> </entry> </row>
</tbody>
</tgroup>
<tfooter>
<SimplePara>My footer</SimplePara>
</tfooter>
</Table>

Thanks.

mrame · November 5th, 2011, 04:13 AM

Its really unclear how your html output should be. Please post the XSLT code tried with and the output html code you are expecting out of the posted XML.

kvbhaskar7 · November 5th, 2011, 05:09 AM

Hi,

Here is the expected output. Also, i need to create the columns dynamically based on the tgroup element <tgroup cols="2">. Also, every <row> <entry align="left" colname="c1"> column name should map to the respective column. Here c1 should go in column1, c2 should display in 2nd column, etc..,

http://i39.tinypic.com/2cpaj9d.jpg

Thanks in advance...

mrame · November 5th, 2011, 08:32 AM

The following code would give you an idea:

Code:

<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
 
<xsl:output method="html"/>
 
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
 
<xsl:template match="thead">
<th>
<xsl:apply-templates select="node()|@*"/>
</th>
</xsl:template>
 
<xsl:template match="row">
<tr>
<xsl:apply-templates select="node()|@*"/>
</tr>
</xsl:template>
 
<xsl:template match="entry">
<td>
<xsl:value-of select="."/>
</td>
</xsl:template>
 
<xsl:template match="tfooter">
<tfoot>
<xsl:value-of select="."/>
</tfoot>
</xsl:template>
 
</xsl:stylesheet>

kvbhaskar7 · November 5th, 2011, 01:37 PM

Thanks for your help.. Will try this code and update..

kvbhaskar7 · November 10th, 2011, 09:36 PM

I found the below solution which is similar to my problem...It worked !

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="TABLE">

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

<xsl:template match="DATAROWS">
<table>
<xsl:apply-templates select="preceding-sibling::HEADERROW" />
<xsl:apply-templates select="DATAROW[position() >= 1]" />
</table>
</xsl:template>

<xsl:template match="HEADERROW | DATAROW">
<tr><xsl:apply-templates /></tr>
</xsl:template>

<xsl:template match="HEADER">
<th><xsl:value-of select="."/></th>
</xsl:template>

<xsl:template match="DATA">
<td><xsl:value-of select="."/></td>
</xsl:template>

</xsl:stylesheet>