Subject: RE: [xsl] Text output: CR LF From: Pieter Reint Siegers Kort <pieter.siegers@xxxxxxxxxxx> Date: Wed, 15 Dec 2004 10:33:11 -0600 |
Hi Jan, I tweaked your XSLT a bit. If you use <xsl:template match="/XML/NODE1"> <xsl:apply-templates select="NODE2" mode="abc"/> </xsl:template> <xsl:template match="NODE2" mode="abc"> <xsl:value-of select="NODE3"/>
 </xsl:template> you get your wanted output. Leaving out '
' gives you your NOT-wanted output. NOTE: This is tested in Xselerator 2.6 using MSXML 4.0. Can't tell for other processors. HTH, <prs/> -----Original Message----- From: news@xxxxxxxxxxx [mailto:news@xxxxxxxxxxx] Sent: Wednesday, December 15, 2004 8:13 AM To: 'xsl-list@xxxxxxxxxxxxxxxxxxxxxx' Subject: [xsl] Text output: CR LF Hello List, I m trying to transform an XML document into plain text using Saxon on a Win XP system. There for I defined a output element: <xsl:output method="text" media-type="text/plain" encoding="iso-8859-1"/> My problem is, that I dont get carriage returns and line feeds where I want them to be (mostly I dont get them at all). For example: <XML> <NODE1> <NODE2> <NODE3>a row</NODE3> </NODE2> <NODE2> <NODE3>a row</NODE3> </NODE2> </NODE1> </XML> I have a template matching to 'NODE1' <xsl:template match="NODE1"> <xsl:apply-templates select="//NODE2" mode="abc"/> </xsl:template> applying over all 'NODE2' nodes. Within this node, I want 'NODE3' as a row. <xsl:template match="NODE2" mode="abc"> <xsl:value-of select="*//NODE3"/>
 </xsl:template> What I get is: a rowarow What I want is: a row a row What are I m doing wrong? Can you help me? Thanks, Jan
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