Hi, I've created a XML data document and an XSLT stylesheet, and I want to output an HTML document based on the two. My stylesheet has the tag, and my XML document has the processor instuction (along with various "xsl:value-of" references). My question is, what are the actual "mechanics" of getting the XSLT processor (which, I understand, is built in to all web browsers) to read the XML and XSLT stylesheet files and output the HTML document, which will then display on the browser? The XML book that I've been reading doesn't specify this! Thank you
My question is, what are the actual "mechanics" of getting the XSLT processor (which, I understand, is built in to all web browsers) to read the XML and XSLT stylesheet files and output the HTML document, which will then display on the browser?
It is the task of the specific HTML browser being used to invoke its XSLT processor. Then the browser interpretes the results of the XSLT transformation as the HTML that should be displayed. Do note that in general browsesers are not required to support XSLT pre-processing, so there may exist browsers that do not have an associated XSLT processor and do not honor the xml-stylesheet
PI for the type="text/xsl"
pseudo-attribute.
For more information read the W3C spec on "Associating Style Sheets with XML Documents"
To test the XSLT transformation in this, somewhat primitive way, you can open the XML file with your browser (do your homework and learn how to open a local file from the browser) and examine the results with a "View Source"
or similar command.
I certainly do not recommend this primitive technique. It is much better to use one of the many existing XSLT IDEs, such as the XSelerator, oXygen, Visual Studio, ..., etc.
You can either run XSL transforms in the "normal way" using Javascript API, or use an xml-stylesheet processing instruction, like this:
Load this in to your browser...
<?xml version="1.0"?>
<?xml-stylesheet href="demo.xslt" type="text/xsl"?>
<data>
<first>first</first>
<second>second</second>
</data>
and the stylesheet, save this as demo.xslt in the same dir as the XML file
<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="html"/>
<xsl:template match="/">
<html>
<head><title>Xslt browser demo</title></head>
<body>
Here's my data:
<xsl:for-each select="/data/*"><b><xsl:value-of select="."/></b></xsl:for-each>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
This works for me in Firefox on Linux.
Dimitre's answer is what you need. But here you have an example:
XML:
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="stylesheet.xsl"?>
<document>
...
</document>
Opening the preceding XML document in any (not really but you get it...) browser and it will transform the XML document with stylesheet.xsl
and display the result.
It's actully quite a mess when it comes to transformations in browsers imo, bad support and only XSLT 1.0. The MIME type See Alejandros comment below.text/xsl
is not even "correct" but it's the one most commonly supported by browsers. The correct MIME type should be application/xslt+xml
but that's not supported in any (?) browser to my knowledge.