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Re: [xsl] xsl transormation from flat tree to hierarchical tree

2012-03-26 10:35:16
Dear Andreas,

First, your code would also be easier to understand if it weren't so verbose. Keep in mind that these two are exactly equivalent:

<xsl:template name="printStateTree"
  xmlns:dotml="http://www.martin-loetzsch.de/DOTML";>
  <xsl:param name="curState"/>
  <xsl:element name="node">
    <xsl:attribute name="id">
      <xsl:value-of select="$curState/@name"/>
    </xsl:attribute>
    <xsl:attribute name="label">
      <xsl:value-of select="$curState/@name"/>
    </xsl:attribute>
    <xsl:attribute name="fontname">Arial</xsl:attribute>
    <xsl:attribute name="fontsize">9</xsl:attribute>
    <xsl:for-each select="//state[@parent = $curState/@name]">
      <xsl:call-template name="printStateTree">
        <xsl:with-param name="curState" select="."/>
      </xsl:call-template>
    </xsl:for-each>
  </xsl:element>
</xsl:template>

<xsl:template name="printStateTree"
  xmlns:dotml="http://www.martin-loetzsch.de/DOTML";>
  <xsl:param name="curState"/>
  <node id="{$curState/@name}" label="{$curState/@name}"
     fontname="Arial" fontsize="9">
    <xsl:for-each select="//state[@parent = $curState/@name]">
      <xsl:call-template name="printStateTree">
        <xsl:with-param name="curState" select="."/>
      </xsl:call-template>
    </xsl:for-each>
  </node>
</xsl:template>

Second, this is easily refactored into templates, which means you don't have to pass a node in for context:

<xsl:template match="state" mode="make-node">
  <node id="{@name}" label="{@name}"
    fontname="Arial" fontsize="9">
    <xsl:apply-templates mode="make-node"
      select="//state[@parent = current()/@name]" />
  </node>
</xsl:template>

(This does the same thing as your code, including using the state/@name twice, once for the node/@id and one for the node/@label. So this will have to be fixed.)

If you don't understand how this works, it's time to study up on templates, template matching and the XSLT processing model.

Cheers,
Wendell

On 3/24/2012 3:15 PM, Mukul Gandhi wrote:
Hi Andreas,
    My apologies for a delayed response (since this question was directed to 
me).

I think, I don't have a concrete suggestion for your new question
below, since it seems to me that this question is very specific to
your application domain.

I'll leave it to anyone else to answer this, or I wish you good luck
to discover the answer yourself.

On Sun, Mar 18, 2012 at 10:30 AM, Andreas Volz<lists(_at_)brachttal(_dot_)net>  
wrote:
Am Sat, 10 Mar 2012 16:00:47 +0530 schrieb Mukul Gandhi:

Hi Mukul,

thanks for that hints! I never used that call-templates before. After
some experiments I got it somehow running in my xsl:

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

<xsl:template name="printStateTree" 
xmlns:dotml="http://www.martin-loetzsch.de/DOTML";>
     <xsl:param name="curState"/>

      <xsl:element name="node">

          <xsl:attribute name="id"><xsl:value-of select="$curState/@name"/>
          </xsl:attribute>

          <xsl:attribute name="label"><xsl:value-of select="$curState/@name"/>
          </xsl:attribute>

          <xsl:attribute name="fontname">Arial</xsl:attribute>

          <xsl:attribute name="fontsize">9</xsl:attribute>

          <xsl:for-each select="//state[@parent = $curState/@name]">

              <xsl:call-template name="printStateTree">
                  <xsl:with-param name="curState" select="."/>
              </xsl:call-template>

         </xsl:for-each>

      </xsl:element>

</xsl:template>

<xsl:template match="/">
    <dotml:graph file-name="stateval" 
xmlns:dotml="http://www.martin-loetzsch.de/DOTML";
      xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance";
      xsi:schemaLocation="http://www.martin-loetzsch.de/DotML ../dotml-1.3/dotml-1.3.xsd" label="stateval" 
fontcolor="#0000A0" fontname="Arial Bold" margin="0.2,0.1" ranksep="0.2" nodesep="0.5"
      bgcolor="#F0F0FF" fontsize="13" style="dashed">

        <xsl:for-each select="stateval/states">

          <xsl:call-template name="printStateTree">
               <xsl:with-param name="curState" select="state[not(@parent)]"/>
          </xsl:call-template>

        </xsl:for-each>

        <xsl:for-each select="stateval/transitions/transition">
            <dotml:edge>
                <xsl:attribute name="from"><xsl:value-of select="@from"/>
                </xsl:attribute>

                <xsl:attribute name="to"><xsl:value-of select="@to"/>
                </xsl:attribute>

                <xsl:attribute name="label"><xsl:value-of select="@event"/>
                </xsl:attribute>

                <xsl:attribute name="fontname">Arial</xsl:attribute>

                <xsl:attribute name="fontsize">7</xsl:attribute>
            </dotml:edge>
        </xsl:for-each>

    </dotml:graph>
</xsl:template>

<xsl:template match="*|text()|@*">
    <xsl:copy>
        <xsl:apply-templates select="@*"/>
        <xsl:apply-templates/>
    </xsl:copy>
</xsl:template>


</xsl:stylesheet>

But the output is still not as I need it:

<?xml version="1.0"?>
<dotml:graph xmlns:dotml="http://www.martin-loetzsch.de/DOTML"; xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"; file-name="stateval" 
xsi:schemaLocation="http://www.martin-loetzsch.de/DotML ../dotml-1.3/dotml-1.3.xsd" label="stateval" fontcolor="#0000A0" fontname="Arial Bold" margin="0.2,0.1" 
ranksep="0.2" nodesep="0.5" bgcolor="#F0F0FF" fontsize="13" style="dashed">
    <node id="Root" label="Root" fontname="Arial" fontsize="9">
        <node id="Initial" label="Initial" fontname="Arial" fontsize="9" />
        <node id="Final" label="Final" fontname="Arial" fontsize="9" />
        <node id="Main" label="Main" fontname="Arial" fontsize="9" />
        <node id="Browser" label="Browser" fontname="Arial" fontsize="9" />
        <node id="Settings" label="Settings" fontname="Arial" fontsize="9" />
        <node id="EMail_Compound" label="EMail_Compound" fontname="Arial" 
fontsize="9">
            <node id="EMail_Compound_Initial" label="EMail_Compound_Initial" 
fontname="Arial" fontsize="9" />
            <node id="EMail_Compound_History" label="EMail_Compound_History" 
fontname="Arial" fontsize="9" />
            <node id="EMail_Read" label="EMail_Read" fontname="Arial" fontsize="9" 
/>
            <node id="EMail_Compose" label="EMail_Compose" fontname="Arial" 
fontsize="9" />
        </node>
    </node>
    <dotml:edge from="Initial" to="Main" label="" fontname="Arial" fontsize="7" 
/>
    <dotml:edge from="Root" to="Final" label="EXIT" fontname="Arial" fontsize="7" 
/>
    <dotml:edge from="Root" to="Main" label="MAIN" fontname="Arial" fontsize="7" 
/>
    <dotml:edge from="Root" to="Browser" label="BROWSER" fontname="Arial" 
fontsize="7" />
    <dotml:edge from="Root" to="Settings" label="SETTINGS" fontname="Arial" 
fontsize="7" />
    <dotml:edge from="Root" to="EMail_Compound" label="EMAIL" fontname="Arial" 
fontsize="7" />
    <dotml:edge from="EMail_Compound" to="EMail_Compound_Initial" label="" fontname="Arial" 
fontsize="7" />
    <dotml:edge from="EMail_Compound_Initial" to="EMail_Compound_History" label="" 
fontname="Arial" fontsize="7" />
    <dotml:edge from="EMail_Compound_History" to="EMail_Read" label="" fontname="Arial" 
fontsize="7" />
    <dotml:edge from="EMail_Compound" to="EMail_Compose" label="EMAIL_COMPOSE" 
fontname="Arial" fontsize="7" />
    <dotml:edge from="EMail_Compound" to="EMail_Read" label="EMAIL_READ" fontname="Arial" 
fontsize="7" />
</dotml:graph>

It has now correct transformation into a parent relation, but dotml needs 
format like this:

<record>
  <node         id="10" label="left"/>
  <node         id="11" label="middle"/>
  <node         id="12" label="right"/>
</record>

<record>
  <node         id="20" label="one"/>
  <node         id="21" label="two"/>
</record>

I tried to learn more about xsl by reading w3schools docu, but I wasn't able to
get exact that result. So I like to have this:

Could you maybe give me please another hint?

regards
  Andreas





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