Oracle XSQL Servlet Parse Error when anchoring a link to an image

Question

I have a database of document links that I wish to portray in a table as graphic images. In other words, each image is a hypertext link to the document.
From my HTML 101 I used to do this as follows:
<IMG SRC="image.gif">
Note, no </IMG> end tag, but this produces an image that I can click to access that document. Lousy form, but it works.
When I run my XSL with corresponding XSQL file to create the HTML file I get the following Servlet Parse Error:
Oracle XSQL Servlet Page Processor 1.0.1.0 (Production)
XSQL-006: XSLT stylesheet is not well-formed: ucmDatabase3.xsl
XML parse error at line 66, char 15
End tag does not match start tag 'img'
I then added the </IMG> end tag,
<IMG SRC="image.gif"> </IMG>
which satisfies the parser, but now I get my graphic image AND the underlined linked text together in the same table cell, which SUCKS...
Is there a way around this madness?
Thanks for your comments..
Jon Kotas
Boeing
Canoga Park, CA

Answer 1

Remember that an XSLT stylesheet is, in particular, a well-formed XML document. In a well-formed XML document, you
cannot have an unterminated element like:
  <foo>
    <bar>  <!-- This elemenent is not closed correctly -->
  </foo>So, in your example, you're running into the problem with
your <IMG> tag, since you have:
  <A>
    <IMG src="...">  <!-- This elt is not closed correctly -->
  </A>You just need to close the IMG tag by changing the closing > to /> instead, like this:
  <A>
    <IMG src="..."/>  <!-- This elt *IS* closed correctly -->
  </A>When it's written out to the browser, you'll see that the
trailing slash will be removed, so the browser will be
happy, too.