Re: Converting linebreak to ?

[Ugo Cei]

Il giorno 27/mag/04, alle 20:42, Stephan Coboos ha scritto:
I'am using a flowscript to retrieve text from a form's textarea. This 
text may contain some linebreaks \n. Now I want display the content 
with line breaks in a html view. So I need to convert \n to <br/>. For 
displaying the content of the textarea I'am using a JXTemplate. How I 
can I transform linebreaks \n into the element <br/>? If i replace all 
\n with <br/> in the html output I will get only &lt;br/&gt;. So I 
have to include the <br/> into the SAX pipeline, but how? Is there a 
easy way to do so (without building a new rome;-)?
  <!--+
      |  Replace newlines with <br>'s
      |  (Credits:  http://www.dpawson.co.uk/xsl/sect2/break.html)
      +-->
        <xsl:template name="substitute">
           <xsl:param name="string" />
           <xsl:param name="from" select="'&#xA;'" />
           <xsl:param name="to">
              <br />
           </xsl:param>
           <xsl:choose>
              <xsl:when test="contains($string, $from)">
                 <xsl:value-of select="substring-before($string, 
$from)" />
                 <xsl:copy-of select="$to" />
                 <xsl:call-template name="substitute">
                    <xsl:with-param name="string"
                                    select="substring-after($string, 
$from)" />
                    <xsl:with-param name="from" select="$from" />
                    <xsl:with-param name="to" select="$to" />
                 </xsl:call-template>
              </xsl:when>
              <xsl:otherwise>
                 <xsl:value-of select="$string" />
              </xsl:otherwise>
           </xsl:choose>
        </xsl:template>

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Ugo Cei - http://beblogging.com/
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