Re: [xsl] Recognizing new line character in XSL

> <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@xxxxxxxxxxxx>
To:         xsl-list@xxxxxxxxxxxxxxxxxxxxxx
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&#xa;
> I am using JDOM as processor and xsl version 1.0.
>
> Thanks much in advance for help.
> Siddhi