Subject: RE: Output a link in a transformation to html
From: Nate Austin <naustin@xxxxxxxxxx>
Date: Thu, 31 May 2001 18:07:58 -0500
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Luare-
XSLT needs to be well-formed XML. I'd suggest doing some reading on exactly
what that means. http://www.w3.org/XML/ should provide you with a few links
for starters. The reason that is not well-formed is because <xsl:value-of/>
is an element. Since HREF is an attribute, it cannot have complex (element)
content. It also must enclose its content in " or '. The solution to your
problem would look something like this:
<h3>Web: <a><xsl:attribute name="href" select="."/><xsl:value-of
select="."/></a></h3>
or (using an Attribute Value Template):
<h3>Web: <a href="{.}"><xsl:value-of select="."/></a></h3>
Hope that helps,
Nate
naustin@xxxxxxxxxx
>Date: Wed, 31 May 2000 21:37:07 +0200
>From: "Luare" <luare@xxxxxxxxxx>
>Subject: Output a link in a transformation to html
>
> Hi,
>
>I am doing a transformation from xml to html.
>This example is not well-formed, why?
>
> <H3>Web: <A HREF= <xsl:value-of select="."/>> <xsl:value-of
select="."/>
></A></H3>
>
>The expresion <xsl:value-of select="."/> return a string, doesn't it?
>
>If I put "<xsl:value-of select="."/>"
>or string(<xsl:value-of select="."/>)
>the error is the same.
>
>It seem a tipical error, but I am a beginner.
>
>Thanks
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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