Hello
I have a problem using XSLT..I want to show my XML file using XSLT.
But, it does not work. Here is my XML file (description2.xml):
<form>
<description>
<desc_id>048/26/SMD/HRD/9/00</desc_id>
<title>Kebijakan2</title>
<dates>3/21/1999</dates>
<author>Smart Co</author>
<file_name>Policy2</file_name>
<type>pdf</type>
<category>memorandum</category>
<sub_cat1>tidak_berlaku</sub_cat1>
<sub_cat2>1983</sub_cat2>
</description>
</form>
Here is my XSL file(description2.xsl):
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:transform version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<html>
<body>
<h2>Document Description</h2>
<table border="1">
<tr bgcolor="#9acd32">
<th align="left">No. Surat</th>
<th align="left">Title</th>
<th align="left">Date</th>
<th align="left">Author</th>
<th align="left">File Name</th>
<th align="left">File Type</th>
<th align="left">Status</th>
<th align="left">Year</th>
</tr>
<xsl:for-each select="form/description">
<tr>
<td><xsl:value-of select="desc_id"/></td>
<td><xsl:value-of select="title"/></td>
<td><xsl:value-of select="dates"/></td>
<td><xsl:value-of select="author"/></td>
<td><xsl:value-of select="file_name"/></td>
<td><xsl:value-of select="type"/></td>
<td><xsl:value-of select="sub_cat1"/></td>
<td><xsl:value-of select="sub_cat2"/></td>
</tr>
</xsl:for-each>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
Here is my ASP code to load XML and XSL(show_xsl.asp):
<html>
<head>
<script>
function loadXMLDoc(fname)
{
var xmlDoc;
// code for IE
if (window.ActiveXObject)
{
xmlDoc=new ActiveXObject("Microsoft.XMLDOM");
}
// code for Mozilla, Firefox, Opera, etc.
else if (document.implementation
&& document.implementation.createDocument)
{
xmlDoc=document.implementation.createDocument(""," ",null);
}
else
{
alert('Your browser cannot handle this script');
}
xmlDoc.async=false;
xmlDoc.load(fname);
return(xmlDoc);
}
function displayResult()
{
xml=loadXMLDoc("description2.xml");
xsl=loadXMLDoc("description2.xsl");
// code for IE
if (window.ActiveXObject)
{
ex=xml.transformNode(xsl);
document.getElementById("example").innerHTML=ex;
}
// code for Mozilla, Firefox, Opera, etc.
else if (document.implementation
&& document.implementation.createDocument)
{
xsltProcessor=new XSLTProcessor();
xsltProcessor.importStylesheet(xsl);
resultDocument = xsltProcessor.transformToFragment(xml,document);
document.getElementById("example").appendChild(res ultDocument);
}
}
</script>
</head>
<body id="example" onLoad="displayResult()">
</body>
</html>
When I run show_xsl.asp , it shows an error..This is an error:
"The stylesheet does not contain a document element. The stylesheet may be empty, or it may not be a well-formed XML document."
Unable to show XML using XSLT
