I have successfully been able to transform a simple xml file with data to another xml file (excel template) using a xsl template, this is what my xsl file looks like:
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" standalone="yes"/>
<xsl:template match="/">
<xsl:processing-instruction name="mso-application">
<xsl:text>progid="Excel.Sheet"</xsl:text>
</xsl:processing-instruction>
...(stuff here)...
</xsl:template>
</xsl:stylesheet>
The resulting xml file is written out correctly BUT with the exception of including
<?xml version="1.0"?>
at the top of the file. How can I get it to appear at the top?
Currently my resulting xml file starts with:
<?mso-application progid="Excel.Sheet"?>
...(rest of file)...
But what I need it to do is:
<?xml version="1.0"?>
<?mso-application progid="Excel.Sheet"?>
.(rest of file)...
I'm doing this transform through a windows form with the following code:
XPathDocument myXPathDoc = new XPathDocument(xmlfile);
XslCompiledTransform myXslTrans = new XslCompiledTransform();
myXslTrans.Load(xslfile);
XmlTextWriter myWriter = new XmlTextWriter(xmlexcelfile, null);
myWriter.Formatting = Formatting.Indented;
myWriter.Namespaces = true;
myXslTrans.Transform(myXPathDoc, null, myWriter);
myWriter.Close();
I've tried playing around with the xsl:output standalone="yes/no"
, as well as omit-xml-declaration="no"
. I've also tried (in the C#) code adding myWriter.WriteStartDocument();
before transforming but that was not allowed. I have tried searching online for this as well and keep coming back to the standalone="yes"
but that isn't working. Is there something I am missing here? Oh and in case you are wondering why I need to have the
<?xml version="1.0"?>
at the top of the resulting file, it's because when opening the xml file with excel, excel doesn't recognize it correctly but if it is included then excel opens it correctly...
The standard way to transform XML data into other formats is by Extensible Stylesheet Language Transformations (XSLT). You can use the built-in XSLTRANSFORM function to convert XML documents into HTML, plain text, or different XML schemas. XSLT uses stylesheets to convert XML into other data formats.
The following is general process to transform the XML document into XSL Formattiong Objects (XSL- FO) and print it. ・ Develop stylesheets that conforms to the DTD of source XML document to create the target output. ・ Input the XML document and the XSLT stylesheet to the XSLT processor to create XSL-FO.
You can do this
<xsl:output method="xml" indent="yes" omit-xml-declaration="no" />
or something similar to this using XmlWriterSettings
Edit: Added more code. Previous was missing some parts
XmlWriterSettings writerSettings = null;
XsltArgumentList transformationArguments = null;
XslCompiledTransform transformer = null;
MemoryStream memoryStream = null;
XPathDocument xPathDocument = null;
StringBuilder sb = null;
XmlWriter writer = null;
XmlDocument resultXml = null;
try
{
writerSettings = new XmlWriterSettings();
writerSettings.OmitXmlDeclaration = false; // This does it
writerSettings.Indent = true;
transformationArguments = new XsltArgumentList();
transformer = new XslCompiledTransform();
memoryStream = new MemoryStream(System.Text.Encoding.Default.GetBytes(xml.OuterXml));
xPathDocument = new XPathDocument(new StreamReader(memoryStream));
sb = new StringBuilder();
// give the settings to the writer here
writer = XmlWriter.Create(sb, writerSettings);
// this is not mandatory, obviously, just passing parameters to my xslt file
foreach (KeyValuePair<string, object> parameter in parameters)
{
transformationArguments.AddParam(parameter.Key, string.Empty, parameter.Value);
}
using (Stream strm = Assembly.GetExecutingAssembly().GetManifestResourceStream("Lib.XSLTFile1.xslt"))
using (XmlReader reader = XmlReader.Create(strm))
{
transformer.Load(reader);
transformer.Transform(xPathDocument, transformationArguments, writer);
}
resultXml = new XmlDocument();
resultXml.LoadXml(sb.ToString());
// for testing only
File.AppendAllText(@"Your path goes here\result.xml", resultXml.OuterXml);
}
catch (Exception)
{
throw;
}
This is how I do it, but this code is specifically written to create an instance of a XmlDocument
. I'm sure you can adapt to your needs.
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