Re: [xsl] Reranging Numbers

[Jeff Sese <jeferson.sese@xxxxxxxxxxxx>]

Works perfectly, thanks!

-- Jeff

On 09 21, 10, at 3:56 PM, Michael Kay wrote:

> First parse the string and convert to a sequence of integers,  
> something like:
>
> <xsl:for-each select="tokenize(., ',\s+')">
> <xsl:choose>
> <xsl:when test="contains(., '-')">
> <xsl:sequence select="xs:integer(substring-before(., '-') to  
> xs:integer(substring-after(., '-')"/>
> </
> <xsl:otherwise>
> <xsl:sequence select="xs:integer(.)"/>
> </
> </
> </
>
> Then do grouping using a method first suggested by David Carlisle on  
> this list:
>
> <xsl:for-each-group select="..." group-adjacent=". - position()">
> <xsl:choose>
> <xsl:when test="count(current-group()) gt 1">
> <xsl:value-of select="current-group()[1], current-group()[last()]"  
> separator="-"/>
> </
> <xsl:otherwise>
> <xsl:value-of select="current-group()[1]"/>
> </
> </
> </
>
> On 21/09/2010 8:09 AM, Jeff Sese wrote:
> > Hi,
> >
> > If i have a node like:
> >
> > <node>1, 2, 3, 5-8, 9, 11, 15</node>
> >
> > How can I make it as:
> >
> > <node>1-3, 5-9, 11, 15</node>
> >
> > I'm using XSLT 2.0
> >
> > Thanks,
> > -- Jeff