RE: [xsl] selecting nodes based on sibling values

Hi Michael.
I don't know (I have so many fast answers already !!) which one is best, but 
what is surprising me is how simple it seems once written...

Thanks a lot
Regards,
Fabien



-----Message d'origine-----
De : Michael Kay [mailto:mike(_at_)saxonica(_dot_)com] 
Envoyé : mardi 23 novembre 2010 16:25
À : xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Objet : Re: [xsl] selecting nodes based on sibling values


<xsl:variable name="gpsizes">
<xsl:for-each-group select="row" group-starting-with="row[level='40']">
<xsl:sequence select="count(current-group()) - 1"/>
</xsl:for-each-group>
</xsl:variable>
<xsl:value-of select="max($gpsizes)"/>

Michael Kay
Saxonica

On 23/11/2010 14:45, Fabien Tillier wrote:

Hi List.
Let's say I have an XML like below...

<data>
      <row>
              <level>40</level>
              <keycode>1254.45.12</keycode>
      </row>
      <row>
              <level>50</level>
              <keycode>1254.45.12.7</keycode>
      </row>
      <row>
              <level>50</level>
              <keycode>1254.45.12.8</keycode>
      </row>
      <row>
              <level>50</level>
              <keycode>1254.45.12.9</keycode>
      </row>

      <row>
              <level>40</level>
              <keycode>99.25.6</keycode>
      </row>
      <row>
              <level>50</level>
              <keycode>99.25.6.45</keycode>
      </row>
      <row>
              <level>50</level>
              <keycode>99.25.6.46</keycode>
      </row>
</data>

What could be the XPath expression to get the maximum number of nodes of
level=50 in data those keycode starts like the level=40 line ?
Here the answer would be 3 as the maximum number of level = 50 nodes for
a given level = 40 is 3
I have tried several things, but I am stuck....
(I am using XSLT 2.0 on Saxon)


Thanks for lending me some neurons...
Best regards,
Fabien



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