RE: [xsl] String manipulation & Line Breaks

["Koes, Derrick" <Derrick.Koes@xxxxxxxxxxxxxxxx>]

I don't think you can use mark-up such as <br/> as the string to replace
&#xA; because it will be treated as a string rather than a node.  Please
explain why I am wrong.



-----Original Message-----
From: Jarno.Elovirta@xxxxxxxxx [mailto:Jarno.Elovirta@xxxxxxxxx] 
Sent: Tuesday, March 25, 2003 10:37 AM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE: [xsl] String manipulation & Line Breaks

Hi,

> I have a need to preserve line breaks in a string of text 
> rendered to html
> output.  I'm not sure exactly how to accomplish this, but was hoping
> something along the lines of the following might work:
> 
> <xsl:call-template name="replace">
> <xsl:with-param name="str" select="bio"/>
> <xsl:with-param name="from">\n</xsl:with-param>
> <xsl:with-param name="to"><br/></xsl:with-param>
> </xsl:call-template>

"\n" would be LINE FEED in e.g. Java, but not in XML.
 
> It doesnt.
> Neither does:
> 
> <xsl:call-template name="replace">
> <xsl:with-param name="str" select="bio"/>
> <xsl:with-param name="from"><xsl:text>
> </xsl:text></xsl:with-param>
> <xsl:with-param name="to"><br/></xsl:with-param>
> </xsl:call-template>

Well, this should work. Are you sure you don't have extra whitespace inside
xsl:text? Anyhow, try

  <xsl:with-param name="from" select="'&#xA;'" />

Cheers,

Jarno - Linnunradan käsikirja liftareille: Osa 14

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