RE: [XSLTalk] Problem with transforming XML to XM.L using XSLT

Hi,
 
Can you tell us how do you exaclty obtain your output? Because I can clearly
see that the desired output is exactly what you get when you perform the
transformation of your sources.... ¿so what's up?
 
Regards, Pieter

-----Original Message-----
From: talktao2003 [mailto:talktao2003@yahoo.com]
Sent: Monday, July 21, 2003 12:34 PM
To: XSLTalk@yahoogroups.com
Subject: [XSLTalk] Problem with transforming XML to XM.L using XSLT


Can anybody let me know why I cannot get the desired output? 
Thanks so much.

INPUT XML FILE:

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="cdcatalog3.xsl"?>
<catalog>
  <cd>
    <name>Name1</name>
    <artist>Bonnie Tyler</artist>
    <company>CBS Records</company>
    <price>9.90</price>
    <year>1988</year>
  </cd>
  <cd>
    <name>Name2</name>
    <artist>Jim Henry</artist>
    <company>London</company>
    <price>7.80</price>
    <year>1987</year>
  </cd>
</catalog>

INPUT XSLT FILE:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" 
xmlns:xsl=" http://www.w3.org/1999/XSL/Transform
<http://www.w3.org/1999/XSL/Transform> ">
<xsl:output method="xml" omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="/">
<catalog>
  <xsl:for-each select="catalog/cd">
    <cd>
      <name>    <xsl:value-of select="name"/>  </name>
      <artist> <xsl:value-of select="artist"/>  </artist>
    </cd>
  </xsl:for-each>
</catalog>
</xsl:template>
</xsl:stylesheet>

DESIRED OUPUT:
<catalog>
  <cd>
    <name>Name1</name>
    <artist>Bonnie Tyler</artist>
  </cd>
  <cd>
    <name>Name2</name>
    <artist>Jim Henry</artist>
  </cd>
</catalog>

OUTPUT GOT USING THE METHOD ABOVE:
Name1 Bonnie Tyler Name2 Jim Henry 



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Received on Monday, 21 July 2003 17:29:06 UTC