RE: Replacing <break> tag when not followed by line fe

Hi,

> I have following xml file
> 	<root>
> 		<long>This is a very <break/>
> 		very long text. 
> 		</long>
> 	<root>
> 
> What I want to do is replace every </break> tag and linefeed 
> by \n\, except
> when the </break> tag is followed by  a linefeed (as in the 
> example). Than I
> only want one \n\ instead of 2. What I have no is a replace 
> function for the
> linefeeds and a template match for the break tags. What I 
> like to do is in
> the template match defining when this node is is followed by 
> a linefeed it
> shouldn't be replaced by \n\. Is this somehow possible ?

<xsl:template match="@*|node()">
  <xsl:copy>
    <xsl:apply-templates select="@*|node()"/>
  </xsl:copy>
</xsl:template>

<xsl:template match="long">
  <xsl:apply-templates select="node()" mode="replace" />
</xsl:template>

<xsl:template match="break" mode="replace">
  <xsl:if test="not(following::node()[1][self::text() and starts-with(., '&#xA;')])">\n\</xsl:if>
</xsl:template>

<xsl:template match="text()" name="replace" mode="replace">
  <xsl:param name="text" select="." />
  <xsl:choose>
    <xsl:when test="contains($text, '&#xA;')">
      <xsl:value-of select="substring-before($text, '&#xA;')" />
      <xsl:text>\n\</xsl:text>
      <xsl:call-template name="replace">
        <xsl:with-param name="text" select="substring-after($text, '&#xA;')" />
      </xsl:call-template>
    </xsl:when>
    <xsl:otherwise>
      <xsl:value-of select="$text" />
    </xsl:otherwise>
  </xsl:choose>
</xsl:template>

Or something in those lines,

--

Jarno - Grendel: Human Saviour

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