TD;LR:
No matter the XSL and XML file, it seems that XSL transformations don't work on Chrome 71 but work on Firefox for instance.
For the sake of example, I have two files (that I borrowed from W3School):
simple.xml:
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="simple.xsl"?>
<catalog>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>1985</year>
</cd>
</catalog>
and simple.xsl:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html" encoding="UTF-8" indent="yes"
doctype-public="-//W3C//DTD HTML 4.01//EN"
doctype-system="http://www.w3.org/TR/html4/strict.dtd"
/>
<xsl:template match="/">
<html>
<body>
<h1>Music Collection:</h1>
<table border="1">
<tr bgcolor="#9acd32">
<th>Title</th>
<th>Artist</th>
</tr>
<tr>
<td><xsl:value-of select="catalog/cd/title" /></td>
<td><xsl:value-of select="catalog/cd/artist" /></td>
</tr>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
When I run on Firefox, it works fine, but on Chrome I have a blank page with this error in the console (that I don't get on Firefox):
Unsafe attempt to load URL file:///[...]/simple.xsl from frame with URL file:///[...]/simple.xml. 'file:' URLs are treated as unique security origins.
Is Chrome even able to run XSL tranformations? I find it strange because the W3School snippets work well on this browser, but when I copy and paste the code and run the code on my computer it doesn't work. What should I do?
Thank you for your help.
