Using the position() function...

vb89 · August 26th, 2009, 06:00 PM

I have the following in an XML i'm working with

snippet

Code:

<conference name="Eastern" id="1"/>
<division name="Northeast" id="1"/>
<opponent-name name="Rockets" alias="Col"/>
<opponent-city city="Colorado"/>
<opponent-code id="17" global-id="4970"/>
<conference name="Western" id="2"/>
<division name="Northwest" id="6"/>

I'n my xsl i've coded the following to retrieve the nodes/attributes and their values...

Code:

 
<xsl:variable name="nm" select="conference/@name"/>
<xsl:variable name="id" select="conference/@id"/>
  <conference name="{$nm}" id="{$id}"/>&newline;
 
<xsl:variable name="nm" select="division/@name"/>
<xsl:variable name="id" select="division/@id"/>
  <division name="{$nm}" id="{$id}"/>&newline;
 
<xsl:variable name="nm" select="opponent-name/@name"/>
<xsl:variable name="alias" select="opponent-name/@alias"/>
  <opponent-name name="{$nm}" alias="{$alias}"/>&newline;
 
<xsl:variable name="city" select="opponent-city/@city"/>
  <opponent-city city="{$city}"/>&newline;
 
<xsl:variable name="id" select="opponent-code/@id"/>
<xsl:variable name="gid" select="opponent-code/@global-id"/>
  <opponent-code id="{$id}" global-id="{$gid}"/>&newline;<xsl:variable name="confnm" select="conference/@name"/>
<xsl:variable name="confid" select="conference/@id"/>
  <conference name="{$confnm}" id="{$confid}"/>&newline;
<xsl:variable name="divnm" select="division/@name"/>
<xsl:variable name="divid" select="division/@id"/>
  <division name="{$divnm}" id="{$divid}"/>&newline;

The problem that i'm currently hacving is that i keep retrieving the same value for the attributes in both the conference and division nodes in the output.

Code:

<!--current output-->
 
<conference name="Eastern" id="1"/>
<division name="Northeast" id="1"/>
<opponent-name name="Rockets" alias="Col"/>
<opponent-city city="Colorado"/>
<opponent-code id="17" global-id="4970"/>
<conference name="Eastern" id="1"/>
<division name="Northeast" id="1"/>
 
<!-- desired output -->
 
<conference name="Eastern" id="1"/>
<division name="Northeast" id="1"/>
<opponent-name name="Rockets" alias="Col"/>
<opponent-city city="Colorado"/>
<opponent-code id="17" global-id="4970"/>
<conference name="Western" id="2"/>
<division name="Northwest" id="6"/>

Is there a way to use the position() function so that i can retrieve the value of the attributes of the second conference and division? I tried things like:

<xsl:variable name="confnm" select="conference/@name[position=last]"/>
<xsl:variable name="confid" select="conference/@id[position=last]"/>
<conference name="{$confnm}" id="{$confid}"/>&newline;

<xsl:variable name="confnm" select="conference/@id[position=2]"/>
<conference name="{$confnm}" id="{$confid}"/>&newline;

But none of the above seems to work.

mhkay · August 26th, 2009, 06:54 PM

You can get the second conference like this

conference[2]/@name

or the last conference like this:

conference[last()]/@name

vb89 · August 26th, 2009, 07:06 PM

Perfect...thanks mhkay!!

vb89 · October 15th, 2009, 11:54 AM

So to follow up on the original question, I have a very similar question but can't seem to get this working...

In the case i have a for-loop and i'm doing something like this:

Code:

<xsl:for-each select="sample/number">
 
               <xsl:choose>
                <xsl:when test="@number=1">
                  <xsl:variable name="xAns" select="@answer"/>
                    <sample-1 answer="{$xAns}"/>&newline;
                </xsl:when>
                </xsl:choose>
 
               <xsl:choose>
                <xsl:when test="@number=2">
                  <xsl:variable name="xAns" select="@answer"/>
                    <sample-2 answer="{$xAns}"/>&newline;
                </xsl:when>
                </xsl:choose>
etc...

So saying the original XML looks something like this....

Code:

<sample number=1 answer=434/>
<sample number=2 answer=438/>
...
<sample number=5 answer=534/>
<sample number=1 answer=538/> and then the numbering restarts back to one

What I would like to know is how to turn that sample number=1 where the numbering RESTARTS to 6. So I would like it to say

Code:

<sample number=5 answer=534/>
<sample number=6 answer=538/>
<sample number=7 answer=542/>
etc...

I was thinking something like

Code:

               <xsl:choose>
                <xsl:when test="@number=1[2]">
                  <xsl:variable name="xAns" select="@answer"/>
                    <sample-6 answer="{$xAns}"/>&newline;
                </xsl:when>
                </xsl:choose>

But the above doens't seem to work.

Martin Honnen · October 15th, 2009, 12:01 PM

Can you post a well-formed XML input sample and the corresponding XML output sample you want to create?
And also mention whether you use XSLT 2.0 or 1.0.

vb89 · October 15th, 2009, 12:16 PM

XSLT: 2.0

Original XML

Code:

<homework>
   <math subjuect=algebra>
     <sample number=1 answer=434/>
     <sample number=2 answer=438/>
     <sample number=3 answer=442/>
     <sample number=4 answer=446/>
     <sample number=5 answer=550/>
     <sample number=1 answer=554/>
     <sample number=2 answer=558/>
     <sample number=3 answer=562/>
     <sample number=4 answer=566/>
     <sample number=5 answer=570/>
  </math>
</homework>

Desired output

Code:

<homework>
   <math subjuect=algebra>
     <sample number=1 answer=434/>
     <sample number=2 answer=438/>
     <sample number=3 answer=442/>
     <sample number=4 answer=446/>
     <sample number=5 answer=550/>
     <sample number=6 answer=554/>
     <sample number=7 answer=558/>
     <sample number=8 answer=562/>
     <sample number=9 answer=566/>
     <sample number=10 answer=570/>
  </math>
</homework>

Martin Honnen · October 15th, 2009, 12:32 PM

You can simply use xsl:number as follows:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="2.0">
  
  <xsl:template match="sample">
    <xsl:copy>
      <xsl:attribute name="number">
        <xsl:number/>
      </xsl:attribute>
      <xsl:attribute name="answer" select="@answer"/>
    </xsl:copy>
  </xsl:template>
  
  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@* | node()"/>
    </xsl:copy>
  </xsl:template>

</xsl:stylesheet>

vb89 · October 15th, 2009, 04:59 PM

Thanks for the reply Martin but I had one more quesitioin. If i wanted to concat a value to the node, how would i go about this?

Desired output

Code:

<homework>
   <math subjuect=algebra>
     <sample-1 number=1 answer=434/>
     <sample-2 number=2 answer=438/>
     <sample-3 number=3 answer=442/>
     <sample-4 number=4 answer=446/>
     <sample-5 number=5 answer=550/>
     <sample-6 number=6 answer=554/>
     <sample-7 number=7 answer=558/>
     <sample-8 number=8 answer=562/>
     <sample-9 number=9 answer=566/>
     <sample-10 number=10 answer=570/>
  </math>
</homework>

mhkay · October 15th, 2009, 05:29 PM

>I can think of any conceivable reason for wanting a sequence of elements with names sample-1, sample-2, etc - it looks like a really bad way of using XML. But you can do it with <xsl:element name="sample-{position()}"/>.

vb89 · October 15th, 2009, 05:47 PM

Yeah i agree this XML is a bit wacky/irregular...regardless you solution did it. Thank you!