Re: [xsl] formatting problem

Subject: Re: [xsl] formatting problem
From: Markus Spath <mspath@xxxxxxxx>
Date: Wed, 20 Feb 2002 21:18:04 +0100


Joerg Heinicke wrote:


Try the following (replacing spaces with non-breaking spaces and linebreaks
with <br/>):

<xsl:template match="ARTICLE">
    <xsl:call-template name="linebreaks">
        <xsl:with-param name="string" select="translate(., ' ', '&#160;')"/>
    </xsl:call-template>
</xsl:template>

<xsl:template name="linebreaks">
    <xsl:param name="string"/>
    <xsl:choose>
        <xsl:when test="contains($string, '&#10;')">
            <xsl:value-of select="substring-before($string, '&#10;')"/>
            <br/>
            <xsl:call-template name="linebreaks">
                <xsl:with-param name="string"
select="substring-after($string, '&#10;')"/>
            </xsl:call-template>
        </xsl:when>
        <xsl:otherwise>
            <xsl:value-of select="$string"/>
        </xsl:otherwise>
    </xsl:choose>
</xsl:template>

This handles only linefeed, maybe you still need to handle carriage return.
we replaced &#13; with 'nothing', so you have to change the with-param in
the first template:

        <xsl:with-param name="string" select="translate(., ' &#13;',
'&#160;')"/>

This means: replace ' ' (space) with &#160; (non-breaking space) and &#13;
(carriage return) with nothing.



or (less sophisticated ;) ):



... <xsl:template match="ARTICLE"> <pre> <xsl:value-of select="."/> </pre> </xsl:template> ...

Markus




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