Re: replacement more than one Tag in node

Hi,
      If you would like to do the replacement on multiple tags, you have
many options.

1. you can do like suggested earlier.

<xsl:template match="name|surname|aaa|bbb">

or

2. Have individual templates for name, surname, aaa and bbb

or

3. or you can use a for-each to iterate over part1 node's child nodes

or

4. do an <xsl:apply-templates select="*"/> inside your template for part1

and do as in 1 or 2 above.

And you don't have to do substring-before/substring-after, you can use the
translate function:

<xsl:value-of select="translate($name, '.', ':')"/>

Also, You don't have a node named ptr1 in your xml.


Cheers,
Omprakash.V









                                                                                                                   
                    henry human                                                                                    
                    <henry_human@        To:     xsl-list@xxxxxxxxxxxxxxxxxxxxxx                                   
                    yahoo.de>            cc:     (bcc: omprakash.v/Polaris)                                        
                                         Subject:       replacement  more than one Tag in  node               
                    03/30/2005                                                                                     
                    03:30 AM                                                                                       
                    Please                                                                                         
                    respond to                                                                                     
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hello,
this time i will to define a template to
replace . with : in more than one Tags
f.e replace in name, surname,aaa,bbb ...
how should the template rule be defined to do this??



<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="my.xsl"?>


<document>

<part1>
<ttt>u.bb</ttt>
<name>u.bb</name>
<surname>u.bb</surname>
<aaa>u.bb</aaa>
<bbb>u.bb</bbb>
</part1>

<part2>

<name>u.bb</name>

</part2>
</document>

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

<xsl:template match="document">

<xsl:apply-templates select="part1"/>
</xsl:template>

<xsl:template match="ptr1">
<xsl:call-template name="part1">
<xsl:with-param name="name" select="name"/>
<xsl:with-param name="search_for" select="'.'"/>
<xsl:with-param name="replace_out" select="':'"/>
</xsl:call-template>
</xsl:template>

 <xsl:template name="part1">


<xsl:param name="name"/>
<xsl:param name="search_for"/>
<xsl:param name="replace_out"/>



<xsl:choose>
<xsl:when test="contains($name,$search_for)">
<xsl:value-of
select="substring-before($name,$search_for)"/>
<xsl:value-of select="$replace_out"/>

<xsl:call-template name="part1">
<xsl:with-param name="name"
select="substring-after($name,$search_for)"/>
<xsl:with-param name="search_for"
select="$search_for"/>
<xsl:with-param name="replace_out"
select="$replace_out"/>
</xsl:call-template>

</xsl:when>

<xsl:otherwise>
<xsl:value-of select="$name"/>
</xsl:otherwise></xsl:choose></xsl:template>
</xsl:stylesheet>






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