RE: Delete XML Node

<X>               maps to---->            <X1>
 <A>A</A>        maps to---->              <A1>A</A>
 <Y>             maps to--->               <Y1>
                 if resultant <Y1>
                 has got elements within it
   <B>B</B>      maps to---->                 <B1>B1</B1>
                 only if data contained
                 in <B> is "ABC"
   <C>C</C>      maps to---->                 <C1>C1</C1>
                 only if data contained
                 in <C> is "XYZ"
  </Y>                                      </Y1>
</X>                                         </X>

Thanks,
Deepak

From: "KIENLE, STEVEN C [IT/0200]" <steven.c.kienle@xxxxxxxxxxxxx>
Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To: "'xsl-list@xxxxxxxxxxxxxxxxxxxxxx'" <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Subject: RE:  Delete XML Node
Date: Thu, 31 Oct 2002 16:14:50 -0500

Replace the for each with the Y[Z] with the following:

  <xsl:for-each select="Y[child::*]">
    <xsl:copy-of select="." />
  </xsl:for-each>

This will fire only for those Y which have children.  Then it makes a deep
copy of that node.

If you need this logic to apply you node more than Y, then you may want to
look at multiple templates instead of for-each elements.

-----Original Message-----
From: Deepak Rao [mailto:deepaksubs@xxxxxxxxxxx]
Sent: Thursday, October 31, 2002 4:04 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE:  Delete XML Node

Whoops, I shot myself in the foot here....

sorry folks I am new to XSL.... and pardon my ignorrance

To clarify things, I want an XSL which would output a <Y> if and only if it
has a sub element in this case a <Z>. In the current example I have only one

sub element <Z> but there can be more than one sub-element of <Y>.

e.g.

<Y>
<Z></Z>
<D></D>
......
</Y>

I do not want <Y></Y> to appear in the output if all the sub-elements do not

map.

Thanks,
Deepak

>From: "KIENLE, STEVEN C [IT/0200]" <steven.c.kienle@xxxxxxxxxxxxx>
>Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
>To: "'xsl-list@xxxxxxxxxxxxxxxxxxxxxx'" <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
>Subject: RE:  Delete XML Node
>Date: Thu, 31 Oct 2002 14:27:42 -0500
>
>Deepak,
>
>In case what you really want to do is output Y if and only if it has a Z
>element node, the following transform will work:
>
><?xml version="1.0"?>
><xsl:stylesheet version="1.0"
>xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
>   <xsl:template match="/X">
>     <X>
>       <xsl:for-each select="A">
>         <A>
>           <xsl:value-of select="."/>
>         </A>
>       </xsl:for-each>
>       <xsl:for-each select="Y[Z]">
>         <Y>
>           <xsl:for-each select="Z">
>             <Z>
>               <xsl:value-of select="."/>
>             </Z>
>           </xsl:for-each>
>         </Y>
>       </xsl:for-each>
>     </X>
>   </xsl:template>
></xsl:stylesheet>
>
>This key is to use the XPath selection of "Y[Z]" which means all Y which
>have a Z element.
>
>With the input XML of:
>
><X>
>   <A>A</A>
>   <Y></Y>
></X>
>
>The output will be:
>
><X>
>   <A>A</A>
></X>
>
>With the input XML of:
>
><X>
>   <A>A</A>
>   <Y>
>     <Z>B</Z>
>   </Y>
></X>
>
>The output will be:
>
><X>
>   <A>A</A>
>   <Y>
>     <Z>B</Z>
>   </Y>
></X>
>
>I hope this may help.
>
>	Steve
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list

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