[xsl] Re: line feeds when outputing as text

["Dimitre Novatchev" <dnovatchev@xxxxxxxxx>]

Use:

     <xsl:value-of select="concat(@url, '&#xA;')"/>


=====
Cheers,

Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL




"Sam Carleton" <sam@xxxxxxxxxxxxxx> wrote in message
news:20030303201331.GA14537@xxxxxxxxxxxxxxxxxxx
> Folks,
>
> I have an xml file that contains item elements.  Some items have an
> url attribute.  I am trying to write a xslt program to give me each
> url on a separate line.  I figured this would work:
>
> <?xml version="1.0" encoding="UTF-8"?>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
>
>   <xsl:output method="text" indent="yes"/>
>
>   <xsl:template match="/">
>     <xsl:apply-templates select=".//item[string-length(@url)!=0]"/>
>   </xsl:template>
>
>   <xsl:template match="item">
>     <xsl:value-of select="@url"/>
>   </xsl:template>
>
> </xsl:stylesheet>
>
> The problem with this is that there is no line breaking between the
> different URL's.  I tried adding both a &#13; and &#10; (forgetting
> which one UNIX likes):
>
>   <xsl:template match="item">
>     <xsl:value-of select="@url"/>&#13;
>   </xsl:template>
>
> And that still had no effect.  Finally I tried to tack something on
> the front and that DID work:
>
>   <xsl:template match="item">
>     :<xsl:value-of select="@url"/>
>   </xsl:template>
>
> I could live with this except for the fact that there is no line on
> the last entry so my shell script is not picking up the last entry.
> How do I get a linefeed at the end?
>
> Sam
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>




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