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Subject: Re: [xsl] Replace newline characters(\n) with <br> From: "Joerg Heinicke" <joerg.heinicke@xxxxxx> Date: Thu, 21 Mar 2002 01:24:00 +0100 |
You can search the archives for 'replace' or 'recursive templates', such
questions are quite often.
<xsl:template match="message">
<xsl:call-template name="replace">
<xsl:with-param name="string" select="."/>
</xsl:call-template>
</xsl:template>
<xsl:template name="replace">
<xsl:param name="string"/>
<xsl:choose>
<xsl:when test="contains($string,' ')">
<xsl:value-of select="substring-before($string,' ')"/>
<br/>
<xsl:call-template name="replace">
<xsl:with-param name="string"
select="substring-after($string,' ')"/>
</xsl.call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$string"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
Regards,
Joerg
> Hello,
>
> I have the a simple Xml document as shown below
>
> <xml version="1.0">
> <message>
> string1
> string2
> string3
> </Message>
>
> Technically, the strings are separated by newlines. Now when i write an
XSL
> for this document, i want to replace all the newlines with <br> so that
> these strings appear in three lines in the output.
>
> Thanks,
>
>
> Arun Ramadoss
> Professional
> Software Engineer II
> E*TRADE Financial
> w 650-331-6307
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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