AW: [xsl] XML Query!

[Delaney Robin <Robin.Delaney@xxxxxxx>]

Hello Rakesh,

I hope not, otherwise I spent a few days wasting my time before Xmas :-)

I had a need also to create CSV from XML and with the help of the many
"unsung heroes" in xsl-list, whose patience and willingness to share
knowledge amazes me at times, I managed to do what I needed. Here's a piece
of what I ended up with. You'll have to change it to suit your needs but it
does the basics. It only extracts the element and attribute values of "leaf
nodes" (descendants that don't have children). The seperator is currently a
"!". 

Hope it helps.

############################
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
	<xsl:output method="text"/>
 	<xsl:template match="xyz">
  		  	<xsl:for-each select="(.//*[not(*)]|.//@*)"
				<xsl:value-of select="."/>	
				<xsl:if test="position()!=last()">!</xsl:if>

			</xsl:for-each> 
			<xsl:text>&#10;</xsl:text>

  	</xsl:template>
    </xsl:stylesheet>
############################


Cheers,

Robin
                                                     
Robin Delaney
FJA Feilmeier & Junker GmbH		    
web  http://www.fja.com

> -----Ursprüngliche Nachricht-----
> Von:	Rakesh Seth [SMTP:RakeshSeth@xxxxxxxxxx]
> Gesendet am:	Freitag, 10. Januar 2003 11:08
> An:	xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Betreff:	[xsl] XML Query!
> 
> Hi
> I want to convert XML data into a csv file.Is there any parser available
> for that?
> 
> Rakesh Sheth
> rakeshseth@xxxxxxxxxx
> rakeshsheth@xxxxxx
> Tel.No.-6952222
> Extn.No.-1122
> Tel.No.(Resi.)-3802293
> 
> 
> 
> 
> MASTEK
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