FF XSL transform returning unexpected result

**acmclellen** · 04-21-2010 01:46 PM

Hi,

I'm currently using the code below to style some XML

Code:

var processor = new XSLTProcessor();
processor.importStylesheet(xslDoc);
var xmlDom = processor.transformToDocument(xmlDoc);
		
var serializer = new XMLSerializer();
output = serializer.serializeToString(xmlDom);

and my stylesheets output method is set to HTML, and it's styling in IE fine, but in Firefox I'm getting the following back:

Code:

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3c.org/tr/html4/strict.dtd">

<transformiix:result xmlns:transformiix="http://www.mozilla.org/TransforMiix">Upcoming Events Thursday, 22 April randal's birthday Please bring cake Thursday, 13 May Ascension  Sunday, 15 August Assumption Day Some description</transformiix:result>

it's not formatted as html at all. Please, any help would be greatly appreciated.

Thank you so much.

**Jona** · 04-21-2010 06:58 PM

Hi,

Can you post an example of what you're working with in a .zip file, please? I'd like to take a look at it, but I don't have the XML or XSL files to do any testing.

**rnd me** · 04-21-2010 09:24 PM

try:

Code:

output=xmlDom.documentElement.innerHTML;

**acmclellen** · 04-23-2010 11:11 AM

Thanks for the response - much appreciated.

I found by changing my output method from

Code:

<xsl:output method="html" />

to

Code:

<xsl:output method="html" indent="yes"
            encoding="iso-8859-1"
            media-type="text/html" doctype-public="-//W3C//DTD HTML 4.0//EN"/>

solved the problem.

Thanks anyways.

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