Re: Transformation from XML to Table Rows

Hi everyone i have been making progress on the XML solution I refered to before. Here is the progress thus far. the everything in the first template match works to perfection. (<xsl:template match="/">)
Within the first template. I'm using a for each instruction. that displays rows.
now within each loop i want it to list all the child sub_topics. 
It's not doing that . if anyone could help with this issue that would be great.
stephen 
once again the xml document structure looks like this.
<root>
<topic id="1">
<topic_name>topic name</topic_name>
<sub_topic id="1">subtopicname</sub_topic>
<sub_topic id="2">subtopicname</sub_topic>
<sub_topic id="3">subtopicname</sub_topic>
</topic>

<topic id="2">
<topic_name>topic name</topic_name>
<sub_topic id="4">subtopicname</sub_topic>
<sub_topic id="5">subtopicname</sub_topic>
<sub_topic id="6">subtopicname</sub_topic>
</topic>
.
.
.
</root>

AND Below is the XSL

  

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html" /> 
<xsl:template match="/">

<BODY bgcolor="white" text="#666666" > 


    <table width="700px" border="0" cellspacing="0" cellpadding="0" align="center"> 
    <tr>
       <xsl:for-each select="channel_list/topic">
 
          <td align="left" valign="absbottom"><span class="channelname">
        <xsl:element name="a">
        <xsl:attribute name="href"> 
            /channel.asp?catid=<xsl:value-of select="@id"/>  
        </xsl:attribute>
            
        <xsl:value-of select="topic_name"/> 
        <hr noshade="noshade" color="#EBEBEB" align="left" width="200px" />
     

            <xsl:if test="position() mod 3 = 0">
                <xsl:element name="tr">
            </xsl:element>
            </xsl:if>
  
        </xsl:element>
        
       </span></td>

    </xsl:for-each>

    </tr>
    </table>
</BODY>
</HTML>

</xsl:template> 



<xsl:template match="topic" mode="sit">
    <table border="0" cellspacing="0" cellpadding="0" align="center"> 
    <tr>
<xsl:for-each select="topic/sub_topic">

<span class="channeltopic">
    <xsl:element name="a">
    <xsl:attribute name="href"> 
        /channel.asp?catid=<xsl:value-of select="@id"/>  
    </xsl:attribute>
    <xsl:value-of select="sub_topic" /> 

    </xsl:element>
</span>
 
</xsl:for-each> 
    </tr>
    </table>
</xsl:template>
</xsl:stylesheet>


In a message dated Mon, 4 Dec 2000  4:53:49 PM Eastern Standard Time, Estefan27@xxxxxxx writes:

<< Greetings;
 
I have an xml document with the following structure: 
<root>
<topic id="1">
<topic_name>topic name</topic_name>
<sub_topic id="1">subtopicname</sub_topic>
<sub_topic id="2">subtopicname</sub_topic>
<sub_topic id="3">subtopicname</sub_topic>
</topic>

<topic id="2">
<topic_name>topic name</topic_name>
<sub_topic id="4">subtopicname</sub_topic>
<sub_topic id="5">subtopicname</sub_topic>
<sub_topic id="6">subtopicname</sub_topic>
</topic>
.
.
.
</root>

Can someone please show me how to make an XSLTransformation that will: 

--Transform the topic into html rows. 3 topics to row.
(i want it flexible so that i can later change the xsl to 4 per row.
therefore i wish to use something like 
    <xsl:if test="position() mod 3 = 0">
        </tr><tr> <!--naturally this gives an error-->
    </xsl:if>
)

all the sub_topics must be presented below each corresponding topic name.
Therefore I'm thinking about something along the lines of the following. (
I could be way off. if someone could help that would be great.
 
<xsl:template match="/">
<table>
<tr>
 <xsl:for-each select="root/topic">
     <td><xsl:apply-templates select="displaytopic" />
     </td>
 <xsl:for-each>
</tr>
</table>
</xsl:template>

<xsl:template match="displaytopic">
  
    <xsl:element name="a">
    <xsl:attribute name="href"> 
    /anasppage.asp?id=<xsl:value-of select="@id"/>  
    </xsl:attribute>
    <xsl:value-of select="topic_name" /> 

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

<xsl:template match="display_related_sub_topic">
  
    <xsl:element name="a">
    <xsl:attribute name="href"> 
    /anasppage.asp?id=<xsl:value-of select="@id"/>  
    </xsl:attribute>
    <xsl:value-of select="sub_topic" /> 

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

all guidance would be greatly appreciated. 

s


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
 >>



 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
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