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| cc: [EMAIL PROTECTED]
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| Subject: Re: XSL example
Hi,
I got XSL working.
Now, it seems that there is some other problem ..
When I try to access an XML page I get an error like
Source XML Error: Value must be quoted.
What does this mean..???
Is there something wrong with the xml file I'm generating..???
I've tried putting values with and
Hi,
May be I should have included my files..
Why do i get the following error while the examples run fine...
Source XML Error: Value must be quoted.
I have tried putting quotes in the xml file, but nothing seems to work.
Pls Help,
Thanks in advance..
Ishpal
I think we would have to see the XML. It appears that an attribute value is
unquoted, though.
Arved Sandstrom
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of Ishpal
Sent: Friday, September 08, 2000 7:58 AM
To: Orion-Interest
Subject: Re: XSL example
The problem is in your XSL file. You have many table elements with unquoted
attribute values.
Kev
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of Ishpal
Sent: 08 September 2000 12:37
To: Orion-Interest
Subject: Re: XSL example
Hi
-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of Ishpal
Sent: 08 September 2000 12:37
To: Orion-Interest
Subject: Re: XSL example
Hi,
May be I should have included my files..
Why do i get the following error while the examples run fine...
Source XML
It's fixed in 1.3.3; thanks!
At 05:06 PM 9/5/00 -0400, you wrote:
How do I run the xsl example in default-web-app/examples/xsl? when using
http://hostname/examples/xsl/doc.jsp I get an error:
Source XML Error: External entity not found:
I don't know if this is a bug, a feature, or if I've just screwed up my
copy of MSIE, but I can't get a formatted web page out of the xsl example
using 1.3.3.
The example doc.jsp links to doc.xsl, which has a output method of xml -
which IE 5.5 displays as xml source. No problem here...
I
We are having an issue with XSL transformations. Here is the following
.jsp we are using:
?xml version="1.0" ?
% int type =