Re: how to replace a part of a string with a newline??

Hi Varun,
 This stylesheet does it. It uses a recursive
template.

<?xml version="1.0"?> <xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">

<xsl:output method="text" />

<xsl:template match="/s">
  <xsl:call-template name="replace-string">
     <xsl:with-param name="str" select="." />
     <xsl:with-param name="delim" select="'__OPT__'"
/>
  <xsl:with-param name="newdelim" select="':'" />
</xsl:call-template>
</xsl:template>

<xsl:template name="replace-string">
<xsl:param name="str" />
<xsl:param name="delim" />
<xsl:param name="newdelim" />
<xsl:if
test="string-length(substring-before($str,$delim)) =
0">
<xsl:value-of select="$str" />
</xsl:if>
<xsl:if
test="string-length(substring-before($str,$delim)) >
0">
<xsl:value-of
select="substring-before($str,$delim)" /><xsl:value-of
select="$newdelim" />
</xsl:if>
<xsl:if
test="string-length(substring-after($str,$delim)) >
0">
<xsl:call-template name="replace-string">
<xsl:with-param name="str"
select="substring-after($str,$delim)" />
<xsl:with-param name="delim" select="$delim"
/>
<xsl:with-param name="newdelim"
select="$newdelim" />
</xsl:call-template>
</xsl:if> </xsl:template>

</xsl:stylesheet>

for e.g. when it is applied to XML
<?xml version="1.0"?>
<s>Importer__OPT__Trading Company__OPT__Service
Provider</s>

it produces output
Importer:Trading Company:Service Provider

Regards,
Mukul

--- Varun Suri <varunsuri@xxxxxxxxxxxxxxx> wrote:

Hi

Ive got this string :

Importer__OPT__Trading Company__OPT__Service
Provider

How do i make it equivalent to
- Importer
- Trading Company
- Service Provider

Ive tried using translate function, but it replaces
all instances of the letters O,P,T as well.I did something like:

translate(text(),'__OPT__',':')

It gives the following output is Importer:rading
Company:Service rovider which i dont want.

Any easy way to do it??

Thanks in advance
Regards
Varun

		
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Thanks Mukul
Worked perfectly.