Error with parameter. Why?

[shapper]

Hello,

Inside my XSL file I have the following:

<xsl:element name="loc">
<xsl:value-of select="$MyParameter" />
<xsl:value-of select="substring(@url, 3)"/>
</xsl:element>

In line <xsl:value-of select="$MyParameter" /> I get an error saying:
The value of parameter MyParameter is either not defined or it is out
of scope.

Thanks,
Miguel

 

[Joseph Kesselman]

Where does the matching xslarameter declaration appear?

 

[Martin Honnen]

Inside my XSL file I have the following:

<xsl:element name="loc">
<xsl:value-of select="$MyParameter" />
<xsl:value-of select="substring(@url, 3)"/>
</xsl:element>

In line <xsl:value-of select="$MyParameter" /> I get an error saying:
The value of parameter MyParameter is either not defined or it is out
of scope.

So where you you have an
<xslaram name="$MyParameter"/>
and where do you set its value (either from the outside if the param is
global or with an <xsl:with-param name="$MyParameter"
select="XPathexpression"/>)?

 

[shapper]

Sorry,

I am lost. This is my first XSL file. Inside my XSL file I have
something like this:

<xsl:element name="loc">
<xsl:text>HERE</xsl:text>
<xsl:value-of select="substring(@url, 3)"/>
</xsl:element>

What I want is to be able to change HERE to something else.
For this I tried to add the MyParameter as I described.

I set the parameter value in my VB.NET code on my Asp.Net web site.
I do that when I run a function to transform the XML file using the XSL
file.

Was this what you asked me?

Sorry, but I am just starting with XSL and XML.

Thanks,
Miguel

 

[Joseph Kesselman]

I set the parameter value in my VB.NET code on my Asp.Net web site.

The parameter's value has to be explicitly passed to your stylesheet
(see the XSLT implementation's documentation for the syntax, since it
varies from processor to processor) and has to be explicitly accepted by
a top-level xslaram element in the stylesheet.

See the XSLT FAQ's section on variables and parameters, and/or find a
good tutorial.