<xsl:apply-template match="text()">
<xsl:for-each select="tokenize(., '\n')">
<p><xsl:value-of select="."/></p>
</
</
Oh, I see you're using XSLT 1.0. That's tedious - in that case you need
...
Or use below slight modification in case the XSLT 1.0 processor supports
EXSLT (http://exslt.org/str/index.html):
<xsl:apply-template match="text()">
<xsl:for-each select="str:tokenize(., '\n')">
<p><xsl:value-of select="."/></p>
</
</
Mit besten Gruessen / Best wishes,
Hermann Stamm-Wilbrandt
Developer, XML Compiler, L3
WebSphere DataPower SOA Appliances
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From: Michael Kay <mike(_at_)saxonica(_dot_)com>
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Date: 07/20/2010 01:33 PM
Subject: Re: [xsl] Recognizing new line character in XSL
Assuming that you apply-templates to text nodes, and that you are
generating HTML, and that you are using XSLT 2.0, you can do
<xsl:apply-template match="text()">
<xsl:for-each select="tokenize(., '\n')">
<p><xsl:value-of select="."/></p>
</
</
Oh, I see you're using XSLT 1.0. That's tedious - in that case you need
a recursive named template that uses substring-before to process the
text before the first newline, and then calls itself passing
substring-after(., '& #xa;') to process the text after the first newline.
Michael Kay
Saxonica
On 20/07/2010 12:13, Siddhi Thakkar wrote:
Hi Experts,
Could you please guide me if there is any way in XSL to recognize new
line character and display each line as a new paragraph? I am not sure of
the Unicode entity of new line char, if we really need it we can assume it
to be

I am using JDOM as processor and xsl version 1.0.
Thanks much in advance for help.
Siddhi
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