Something like this could be used
<xsl:template name="replace">
<xsl:param name="string" select="." />
<xsl:choose>
<xsl:when test="not($string)" />
<xsl:when test="contains($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" />
<br />
</xsl:otherwise>
</xsl:choose>
</xsl:template>
<xsl:template match="string">
<tr>
<td>
<xsl:call-template name="replace" />
</td>
</tr>
</xsl:template>
http://aspn.activestate.com/ASPN/Mail/Message/xsl-list/1393343
----- Original Message -----
From: "Ben Trafford" <ben(_at_)prodigal(_dot_)ca>
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Tuesday, December 02, 2003 2:08 PM
Subject: [xsl] Line end conversion.
Hi!
I have an odd little problem. I want to match all linefeeds with a given
element's content, and convert them to "\n". Does anybody know of a good
way to do so?
In other words...
<element>This is a phrase with a line end.
It's very frustrating.</element>
...becomes...
<element>This is a phrase with a line end.\n\nIt's very
frustrating.</element>
--->Ben
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list