Re: newbie variable question

Subject: Re: newbie variable question
From: "John E. Simpson" <simpson@xxxxxxxxxxx>
Date: Tue, 26 Sep 2000 14:14:53 -0400
At 12:01 PM 09/26/2000 -0400, Les Howard wrote:
if there are any grp/type element whose contents are "DMA" I want to
print out the value of the associated name in that grp.  if there is no
grp/type element whose contents is "DMA" I want to print out the values
of the description element.

  <xsl:for-each select="area/grp">
    <xsl:if test="string(type)='DMA'">
      <xsl:value-of select="name"/>
     <xsl:variable name="ValueExists" select="'True'"/>
<xsl:if  test="$ValueExists!="'True'">
 <xsl:value-of select="description" />

but I keep getting an error:

A reference to variable or parameter 'ValueExists' cannot be resolved.
The variable or parameter may not be defined, or it may not be in scope.

The error makes sense. XSLT variables are in effect ("in scope") only within the block of code in which they're defined. Your stylesheet is structured in part like this:

   Begin for-each....
     If some condition....
       Define variable $ValueExists
     End if
   End for-each

   If some condition....
     Do something with $ValueExists
   End if

See? The if-block doesn't know anything about $ValueExists. The second, standalone if-block can't "see into" the if-block which contains the variable's definition; the variable goes out of scope when that if-block closes.

You can sometimes get around the problem by declaring the variable globally -- that is, make the <xsl:variable...> element a top-level element (child of <xsl:stylesheet>). I say sometimes because you can't always get at some information that would be readily available from within a template rule.

Might be better in your case to simply instantiate one template for type="DMA" and a different one for type!="DMA", using those expressions as predicates.

John E. Simpson | "If you were going to | shoot a mime, would you use
XML Q&A: | a silencer?" (Steven Wright)

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