xsl:variable assignment clarification

Question

My understanding is that <xsl:variable> is immutable and cannot be reassigned.

I am new to XSL and came across a situation like what is in the example below.

<xsl:stylesheet>

  <xsl:variable name="temp" select="true()"/>

  <xsl:template name="example">
     <xsl:variable name="temp" select="false()"/>
     <p><xsl:value-of select="$temp"/></p>
  </xsl:template>

</styleheet>

I have not found anything definitive as to why this occurs. The only way I can reason that I'm not getting an error and why temp will output false is that there is a global temp variable AND an a local temp variable (and somehow are not colliding).

Why am I able to "reassign" temp?

The second one is local to the template. You're not overriding it in the true sense, but you are in effect making the outer version inaccessible. [Fiddle](www.xmlplayground.xyz/9Usr9c) — Mitya
– Mitya, Commented Feb 15, 2019 at 21:13

michael.hor257k · Accepted Answer · 2019-02-15 21:18:04Z

1

You are able to "reassign" (more precisely, to shadow) the variable, because the first binding is at the top-level of the stylesheet, while the second one is in a template.

From XSLT 1.0 specification:

A binding shadows another binding if the binding occurs at a point where the other binding is visible, and the bindings have the same name. It is an error if a binding established by an xsl:variable or xsl:param element within a template shadows another binding established by an xsl:variable or xsl:param element also within the template. It is not an error if a binding established by an xsl:variable or xsl:param element in a template shadows another binding established by an xsl:variable or xsl:param top-level element.

answered Feb 15, 2019 at 21:18

michael.hor257k

118k6 gold badges36 silver badges55 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

zx485 · Accepted Answer · 2019-02-15 21:32:34Z

1

My understanding is that is immutable and cannot be reassigned.

That assumption is correct. XSLT is a functional language and immutable variables are common in this type of language.

Why am I able to "reassign" temp?

The reason that happens is defined by the term Scope. This means that your second definition of the variable temp overrides/overlays your first one - unless you leave its scope which is - here - the template. But here (as mentioned by @michael.hor257k in the comments) the scope can only be stylesheet-wide or template-wide - so a redefinition of the variable in the same xsl:template - even in another code block - is prohibited.

The only way I can reason that I'm not getting an error and why temp will output false is that there is a global temp variable AND an a local temp variable

In fact, both variables are local, but on another level. The first definition is at the xsl:stylesheet level and the second one at the xsl:template level. The first one may be considered a global variable, but that's only a matter of definition.

edited Feb 15, 2019 at 21:32

answered Feb 15, 2019 at 21:16

zx485

29.1k28 gold badges55 silver badges65 bronze badges

1 Comment

michael.hor257k Over a year ago

That is not entirely correct. It is not only a matter of scope. You're not allowed to override a variable defined in the same template, even for a limited scope.

Collectives™ on Stack Overflow

xsl:variable assignment clarification

2 Answers 2

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related