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re: general problem
- From: saodl at brsi dot org
- To: andrew at thebristoldirectory dot com
- Cc: xsl-list at lists dot mulberrytech dot com
- Date: Fri, 18 Jan 2002 08:52:00 -0800 (PST)
- Subject: [xsl] re: general problem
- Reply-to: xsl-list at lists dot mulberrytech dot com
i'm no javascript expert, but i'm using a piece of
code like this to grab the current xml document for
manipulation in a javascript function.
var xmlDoc = document.XMLDocument;
if (xmlDoc == null){
xmlDoc = navigator.XMLDocument;
}
hope that helps.
siri
------andrew@thebristoldirectory.com wrote------
Hi all,
Imagine I have a set of nodes like this:
<root>
<a>
<b>
<c>
<d>
<e>
</a>
<a>
<b>
<c>
<d>
<e>
</a>
</root>
I apply a stylesheet to the xml to output a table that
displays certain
elements of each <a> such as:
<tr>
<td>b , c, d</td>
</tr>
Say I want to include a button next to that row that,
when clicked, opens a
window displaying -all- the elements for that <a>.
I can do this by calling a javascript function, passing
the position of the
<a> node as a parameter, and using the DOM to extract
the information I
want. However, this currently relies on me hard-coding
the name of the xml
file (xmlObj.load('xmlfile')) into the function, which
then makes the
stylesheet useless for all but one xmlfile.
If you understand what Im asking ;) is there a better
way of doing this?
Thanks to anyone on my wavelength, apolgies to everyone
else
cheers
andrew
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