Sort with XSL

crazeydazey · November 17th, 2009, 06:03 AM

Hi,

I have a strange question which I hope someone will be help me with...

if I had some XML like below..

Code:

<children>
 <child name="bob" />
 <child name="dave" />
 <child name="sid" seq="3" /> 
 <child name="fred" seq="1" />
 <child name="rupert" seq="2"/>
 <child name="mildred" />
</children>

and I wanted to sort ONLY the children that have a seq atrribute all the other children MUST stay in the position they are in.. (as below)

Code:

<children>
 <child name="bob" />
 <child name="dave" />
 <child name="fred" seq="1" /> 
 <child name="rupert" seq="2"/>
 <child name="sid" seq="3" />
 <child name="mildred" />
</children>

is this possible??

I know you can sort with XSL using the xsl:sort element, but will it work in this case??

Cheers

Darren

mhkay · November 17th, 2009, 06:09 AM

I don't think the problem is well specified. In your example, the elements with a seq attribute are in a single contiguous group. What output would you want if that were not the case?

Also you don't say whether you are sorting by seq or by name - in your example the answer is the same either way.

If you want to sort groups of contiguous elements having a seq attribute, you could use

Code:

<xsl:for-each-group select="child" group-adjacent="boolean(@seq)">
  <xsl:choose>
     <xsl:when test="current-grouping-key()">
        <xsl:perform-sort select="current-group()">
            <xsl:sort select="@seq"/>
        </xsl:perform-sort>
     </xsl:when>
     <xsl:otherwise>
         <xsl:sequence select="current-group()"/>
     </
  </
</

crazeydazey · November 17th, 2009, 06:15 AM

yeah sorry about me not been very clear about what I mean, it's a little hard to explain exactly what I do mean..

basic I only want to sort nodes that have a seq attribute and I want to sort them by seq, if they don't have a seq attrib it is essential that they stay where they are..

I hope this clears things up a little..

samjudson · November 17th, 2009, 06:22 AM

It also isn't clear from your example whether the @seq attribute will always be contiguous and consecutive , i.e. will there always be elements which each number in the sequence, or might there be element with @seq = 1, 2, 4 and 5, with no 3.

Code:

   <xsl:template match="child[@seq]">
      <xsl:variable name="seq">
         <xsl:number count="//child[@seq]"/>
      </xsl:variable>
      <child name="{../child[@seq=$seq]/@name}" seq="{$seq}"/>
   </xsl:template>
   <xsl:template match="child[not(@seq)]">
      <xsl:copy-of select="."/>
   </xsl:template>

mhkay · November 17th, 2009, 06:23 AM

>I hope this clears things up a little..

No, it doesn't clear things up at all.

What output would you want for this input?

Code:

<children>
 <child name="bob" seq="3"/>
 <child name="dave" />
 <child name="sid" /> 
 <child name="fred" seq="1" />
 <child name="rupert" seq="2"/>
 <child name="mildred" />
</children>

crazeydazey · November 17th, 2009, 06:34 AM

I kinda wish I'd not started this now, I knew it wasn't gonna be easy for me to explain..

@mhkay from your example I would want..

Code:

<children>
<child name="dave" />
<child name="sid" /> 
<child name="fred" seq="1" />
<child name="rupert" seq="2"/>
<child name="bob" seq="3"/>
<child name="mildred" />
</children>

@samjudson - no it won't always be contiguous and consecutive, it could go ..

Code:

<children>
 <child name="bob" seq="5"/>
 <child name="dave" />
 <child name="sid" /> 
 <child name="fred" seq="3" />
 <child name="rupert" seq="2"/>
 <child name="mildred" />
</children>

samjudson · November 17th, 2009, 06:40 AM

From Michael's example - how do you determine where the moved children now come - they appear in your output grouped together, but with no indication of why they appear after sid and before mildred.

mhkay · November 17th, 2009, 06:40 AM

I don't think we can code this one until we have a clear unambiguous specification.

crazeydazey · November 17th, 2009, 06:49 AM

not to worry guys, there is a hell of a lot more to the full story, which I couldn't begin to explain..

I will have to try and think of some other way around this (ie making sure the XML comes out in the order I want in the first place, I think this would be easier for me.. XSLT is NOT my strong point..)

thanks for your help and time though.. much appreciated...

mhkay · November 17th, 2009, 06:58 AM

>XSLT is NOT my strong point

We're here to help with the XSLT. But we can't help you specify the problem.