[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: Replace newline characters(\n) with <br>
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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