Re: how to replace line-feeds with space?

Chad Small wrote:
> I found some info on dpawson's basic trouble-shooting under the Replace
> link, but can't seem to make the transformation work.

Actually, you're doing way too much work. If you're replacing a single
character with a single character, all you need is the XPath function
translate().

   <xsl:template match="TITLE/text()">
     <xsl:value-of select="translate(.,'&#10;',' ')"/>
   </xsl:template>

Using tail recursion and substrings is only necessary if you're 
doing substring replacement, or replacing characters with elements.

   - Mike
____________________________________________________________________
Mike J. Brown, software engineer at         My XML/XSL resources:
webb.net in Denver, Colorado, USA           http://www.skew.org/xml/


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list