Re: [xsl] Adding element to xml snippet using xslt

Subject: Re: [xsl] Adding element to xml snippet using xslt
From: Nishi Bhonsle <nishi.bhonsle@xxxxxxxxxx>
Date: Tue, 22 Feb 2005 11:25:45 -0800
Joris Gillis wrote:

Tempore 23:01:31, die 02/21/2005 AD, hinc in xsl-list@xxxxxxxxxxxxxxxxxxxxxx scripsit Nishi Bhonsle <nishi.bhonsle@xxxxxxxxxx>:

<xsl:template match="server">
    <xsl:element name="drillOutDir">
        <xsl:value-of select="abc:configuration/drillOutDir"/
    </xsl:element>
    <xsl:element name="started">
        <xsl:copy-of select="@status[. = 'true' ]"/>
        <xsl:value-of select="../abc:configuration/started"/>
    </xsl:element>
</xsl:template>


<xsl:copy-of select="../abc:configuration/started/@status[. = 'true' ]"/>


But I actually don't understand why you create the 'drillOutDir' and 'started' elements inside a template that matches 'server' elements...

Hi:
I tried the above select statement in the xslt, but it still yields an empty 'started' element in the output file.
ie
<xsl:element name="started">
<xsl:copy-of select="../abc:configuration/started/@status[. = 'true' ]"/>
</xsl:element>


If I create the 'drillOutDir' and the 'started' elements outside the 'server' elements template then I noticed that after applying the transformation, the resultant output does not contain them AT ALL.

I tried the below, but neither worked. All the below templates result in NOT containing the 'drillOutDir' and 'started' nodes in the output file.

<xsl:template match="abc:configuration/drillOutDir">
  <xsl:element name="drillOutDir">
     <xsl:value-of select="abc:configuration/drillOutDir"/>
  </xsl:element>
  </xsl:template>

<xsl:template match="abc:configuration">
  <xsl:element name="drillOutDir">
     <xsl:value-of select="abc:configuration/drillOutDir"/>
  </xsl:element>
  </xsl:template>

Can you suggest anything?

Thanks.

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