[xsl] Re: replacing parts of a string

["Dimitre Novatchev" <dnovatchev@xxxxxxxxx>]

"Jonny Pony" <jonnypony666@xxxxxxxxxxx> wrote in message
news:Law14-F78xpTljBOh8s0000768c@xxxxxxxxxxxxxx
> Hi,
>
> forgot the previous mail:
>
> How and where must I customize your template to pass only a string.
> Something like:
>
> <xsl:call-template name="LongString">
> <xsl:with-param name="stringIn"><xsl:value-of
select="."/></xsl:with-param>
> </xsl:call-template>
>
> Thanks
> Jonny

You already have this in the example:

       <xsl:call-template name="str-map">
         <xsl:with-param name="pFun" select="$vTestMap"/>
         <xsl:with-param name="pStr" select="."/>
       </xsl:call-template>

So, just specify the value of the "pStr" parameter, e.g.:

         <xsl:with-param name="pStr"
          select="76rt,dsfg,dsfg"/>


Dimitre Novatchev.
FXSL developer,

http://fxsl.sourceforge.net/ -- the home of FXSL
Resume: http://fxsl.sf.net/DNovatchev/Resume/Res.html




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