Re: [xsl] replacement example , help

[henry human <henry_human@xxxxxxxx>]

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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