RE: [xsl] replacement example , help

["Michael Kay" <mike@xxxxxxxxxxxx>]

You misunderstood the message and have integrated the new code incorrectly.

You need your original named template, which does the string replacement,
and you need code that invokes it when you get to a suitable place in the
source document. You also need to supply values for the parameters. So:

> <xsl:stylesheet version="1.0"
>   xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
> <xsl:output method="html"/> 
> 
> <xsl:template match="/">
>   <xsl:apply-templates/>
> </xsl:template>

  <xsl:template match="document">
    <xsl:copy><xsl:apply-templates/></xsl:copy>
  </xsl:template>
  
  <xsl:template match="part1">
    <xsl:call-template name="part1">
      <xsl:with-param name="name" select="name"/>
      <xsl:with-param name="aa" select="'aa'"/>
      <xsl:with-param name="cc" select="'cc'"/>
    </xsl:call-template>
  </xsl:template>

  <xsl:template name="part1">
    ... your original code...
  </xsl:template>

  </xsl:stylesheet>

Michael Kay
http://www.saxonica.com/


> -----Original Message-----
> From: henry human [mailto:henry_human@xxxxxxxx] 
> Sent: 29 March 2005 13:37
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re: [xsl] replacement example , help
> 
> i still becomm error in IE: part1 template has'nt
> exist.
> ??
> here new code as you said:
> 
> <xsl:stylesheet version="1.0"
>   xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
> <xsl:output method="html"/> 
> 
> <xsl:template match="/">
> <xsl:apply-templates select="part1"/>
> </xsl:template>
> 
> <xsl:template match="part1">
> <xsl:param name="name"/>
> <xsl:param name="aa"/>
> <xsl:param name="cc"/>
> <xsl:choose>
> <xsl:when test="contains($name,$aa)">
> <xsl:value-of select="substring-before($name,$aa)"/>
> <xsl:value-of select="$cc"/>
> 
> <xsl:call-template name="part1">
> <xsl:with-param name="name"
> select="substring-after($name,$aa)"/>
> <xsl:with-param name="aa" select="$aa"/>
> <xsl:with-param name="cc" select="$cc"/>
> </xsl:call-template>
> 
> </xsl:when>
> 
> <xsl:otherwise>
> <xsl:value-of select="$name"/>
> </xsl:otherwise>
> 
> </xsl:choose>
> 
> </xsl:template>
> 
> </xsl:stylesheet>
> 
> 
> 
> 
> 
> 
> 
>  --- omprakash.v@xxxxxxxxxxxxx schrieb: 
> > 
> > 
> > Hi,
> >          You have a named template and you have the
> > call to the template
> > inside the named template.
> > 
> > What you shoud be doing is something like:
> > 
> > <xsl:template match="/">
> >      <xsl:apply-templates/>
> > </xsl:template>
> > 
> > <xsl:template match="part1">
> > <xsl:call-template name="part1"/>
> >      <xsl:with-param ... etc
> > </xsl:template>
> > 
> > You may want to rename your named-template to
> > something other than part1
> > though.
> > 
> > Cheers,
> > Omprakash.V
> > 
> > 
> > 
> > 
> >                                                     
> >                                                     
> >          
> >                     henry human                     
> >                                                     
> >          
> >                     <henry_human@        To:    
> > xsl-list@xxxxxxxxxxxxxxxxxxxxxx                     
> >              
> >                     yahoo.de>            cc:    
> > (bcc: omprakash.v/Polaris)                          
> >              
> >                                          Subject:   
> >  [xsl]  replacement example , help                  
> >          
> >                     03/29/2005                      
> >                                                     
> >          
> >                     04:31 PM                        
> >                                                     
> >          
> >                     Please                          
> >                                                     
> >          
> >                     respond to                      
> >                                                     
> >          
> >                     xsl-list                        
> >                                                     
> >          
> >                                                     
> >                                                     
> >          
> >                                                     
> >                                                     
> >          
> > 
> > 
> > 
> > 
> > hello,
> > In this xsl styesheet i will replace aa with the
> > string cc,
> > what do i wrong ,that
> > i dont get cc?
> > thank you to have a look on this stylesheet:
> > 
> > 
> > <?xml version="1.0"?>
> > <?xml-stylesheet type="text/xsl"
> > href="replace.xsl"?>
> > 
> > 
> > <document>
> > <part1>
> > <name>uaabb</name>
> > </part1>
> > </document>
> > 
> > ------------
> > 
> > 
> > 
> > <xsl:stylesheet version="1.0"
> >   xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
> > <xsl:output method="html"/>
> > 
> > <xsl:template name="part1">
> > <xsl:param name="name"/>
> > <xsl:param name="aa"/>
> > <xsl:param name="cc"/>
> > <xsl:choose>
> > <xsl:when test="contains($name,$aa)">
> > <xsl:value-of select="substring-before($name,$aa)"/>
> > <xsl:value-of select="$cc"/>
> > 
> > <xsl:call-template name="part1">
> > <xsl:with-param name="name"
> > select="substring-after($name,$aa)"/>
> > <xsl:with-param name="aa" select="$aa"/>
> > <xsl:with-param name="cc" select="$cc"/>
> > </xsl:call-template>
> > 
> > </xsl:when>
> > 
> > <xsl:otherwise>
> > <xsl:value-of select="$name"/>
> > </xsl:otherwise>
> > 
> > </xsl:choose>
> > 
> > </xsl:template>
> > 
> > </xsl:stylesheet>
> > 
> > 
> > 
> > 
> > 
> > 
> > 
> >
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