Re: [xsl] problem with Passing Parameters to Templates

Subject: Re: [xsl] problem with Passing Parameters to Templates
From: "helen li" <helen_p_li@xxxxxxxxxxx>
Date: Mon, 22 Jan 2001 10:09:48 -0500
Also I have tried to use the following code to find and give the value of the element that is a descendant of the current node that is called 'codec'. It does not work either.

 <xsl:template name="opt_template">
     <xsl:param name="node"></xsl:param>
          <xsl:variable name="node-element"
                   select=".//*[local-name() = $node]" />
         <xsl:if test="$node-element">
           <xsl:value-of select="$node-element" />
         </xsl:if>
  </xsl:template>

Thanks,

Helen

From: "helen li" <helen_p_li@xxxxxxxxxxx>
Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] problem with Passing Parameters to Templates
Date: Mon, 22 Jan 2001 09:43:41 -0500

Hi Jeni&David,

Thanks for your help.

What I really want to do is to output the value of element codec(not the
value of variable $node) if the element codec exists.  With the following
code, I did not get the result as expected. Did I do something wrong?

 <xsl:template name="opt_template">
     <xsl:param name="node"></xsl:param>
         <xsl:if test=".//*[$node]">
           <xsl:value-of select=".//*[name()=$node]"/>
         </xsl:if>
 </xsl:template>

 <xsl:call-template name="opt_template">
          <xsl:with-param name="node" select="'codec'"/>
 </xsl:call-template>

Thanks again,

Helen

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