I am converting xml to html using xslt 3.0 saxon-HE 9.8 library. Using it in c# code.
I am passing xml and xslt file path in input to get it transformed and get output.
Can anyone please let me know how can I pass xml as string and xslt as string as input in c# code for processing it.
Below is my code.
public static string Transform_XML(string param, string inputfile, string xsltfilename)
{
var xslt = new FileInfo(xsltfilename);
var input = new FileInfo(inputfile);
// Compile stylesheet
var processor = new Processor();
var compiler = processor.NewXsltCompiler();
var executable = compiler.Compile(new Uri(xslt.FullName));
XPathDocument doc = new XPathDocument(new StringReader(param));
DocumentBuilder db = processor.NewDocumentBuilder();
XdmNode xml;
using (XmlReader xr = XmlReader.Create(new StringReader(param)))
{
xml = db.Build(xr);
}
// Do transformation to a destination
var destination = new DomDestination();
using (var inputStream = input.OpenRead())
{
var transformer = executable.Load();
transformer.SetParameter(new QName("", "", "user_entry"), xml);
transformer.SetInputStream(inputStream, new Uri(input.DirectoryName));
transformer.Run(destination);
}
return destination.XmlDocument.InnerXml.ToString();
}
Want to pass xml and xslt as string instead of file path.
UPDATE 1
Got the solution for passing xml and xsl as string in c#. Below is the updated code.
private string Transform_XML(string param, string param_name, string inputfile, string xsltfilename)
{
string xslt_input = System.IO.File.ReadAllText(xsltfilename + ".xslt");
string xml_input = System.IO.File.ReadAllText(inputfile + ".xml");
// Compile stylesheet
var processor = new Processor();
var compiler = processor.NewXsltCompiler();
compiler.BaseUri=new Uri(Server.MapPath("/"));
var executable = compiler.Compile(new XmlTextReader(new StringReader(xslt_input)));
XPathDocument doc = new XPathDocument(new StringReader(param));
DocumentBuilder db = processor.NewDocumentBuilder();
XdmNode xml;
using (XmlReader xr = XmlReader.Create(new StringReader(param)))
{
xml = db.Build(xr);
}
//xml input
DocumentBuilder builder = processor.NewDocumentBuilder();
builder.BaseUri= new Uri(Server.MapPath("/"));
MemoryStream ms = new MemoryStream();
StreamWriter tw = new StreamWriter(ms);
tw.Write(xml_input);
tw.Flush();
Stream instr = new MemoryStream(ms.GetBuffer(), 0, (int)ms.Length);
XdmNode input = builder.Build(instr);
// Do transformation to a destination
var destination = new DomDestination();
var transformer = executable.Load();
//Set the parameter with xml value
transformer.SetParameter(new QName("", "", param_name), xml);
// Set the root node of the source document to be the initial context node
transformer.InitialContextNode = input;
transformer.Run(destination);
// Get result
return destination.XmlDocument.InnerXml.ToString();
}
The XsltTransformer has a method SetInputStream() that allows you to supply the input as a stream (which indeed you appear to be using).
This post How do I generate a stream from a string? tells you how to create a stream from a string.
I want to check if my file does exist first, but if not.. How can i make it exist?
if (File.Exists(Filepath))
{
// if it does exist, itll show datas as data grid view
dataGridView2.DataSource = ds.Tables[0].DefaultView;
}
else
{
// if it doesnt exist, how can i make it exist? or create an XML file
}
You could use XmlDocument and XmlTextWriter
XmlDocument doc = new XmlDocument();
doc.LoadXml("<sample></sample>"); //your content here
// Save the document to a file
XmlTextWriter writer = new XmlTextWriter("sample.xml", null);
doc.Save(writer);
You could also just use a FileStream with the File class.
if (!File.Exists(Filepath))
{
using (FileStream fs = File.Create(Filepath))
{
Byte[] info = new UTF8Encoding(true).GetBytes("Text in the file.");
// Add some information to the file.
fs.Write(info, 0, info.Length);
}
}
Another approach by using LINQ to XML types without explicit using of writers
if (File.Exists(Filepath))
{
// do something
}
else
{
var document =
new XDocument(new XElement("root",
new XElement("one", "value 1"),
new XElement("two", "value 2"));
document.Save(FilePath);
}
using System.Xml.Linq;
XDocument doc = new XDocument(
new XElement("YourNodeName")
);
doc.Save("your_doc_name.xml");
Use StreamWriter to create a new file, specify the path.
You can do this:
using(var tw = new StreamWriter(path, true))
{
tw.WriteLine("New file content");
}
How do I obtain the Diffgram of a DataSet in an XElement? (Or XDocument)
I found out how to obtain the Diffgram in a string:
// DataSet to Diffgram in a string:
StringBuilder sb = new StringBuilder();
XmlWriterSettings settings = new XmlWriterSettings { Indent = true, Encoding = Encoding.UTF8 };
using (XmlWriter xw = XmlWriter.Create(sb, settings))
{
ds.WriteXml(xw, XmlWriteMode.WriteSchema);//XmlWriteMode.DiffGram
}
string str = sb.ToString();
but it seems wasteful to first output the xml of a diffgram to a string and then parse it back in to an XElement. So I am trying to find out how to fill in the missing link in this code, which should transfer the xml of the diffgram without conversions to an Xml variable:
// I have a DataSet filled with some data
DataSet ds = new DataSet();
ds.Tables.Add(ScanData);
// I need the diffgram of the DataSet in an XElement
XElement xe = null;
XmlReader xr = xe.CreateReader();
// I could live with output to XDocument, and extract the XElement later
XDocument xd = new XDocument();
XmlReader xrd = xd.CreateReader();
// Q: How do I construct a stream that connects an XmlReader to ds.WriteXml()?
Stream stream = ...???... ;
// This method creates the DiffGram output format to a stream
ds.WriteXml(stream, XmlWriteMode.WriteSchema);//diffgram output
I hope to find the answer to my code problem, and maybe even to learn how stream/reader/writer really work.
You need a writer, not reader, to write from DataSet to XDocument:
XDocument xd = new XDocument();
using (var writer = xd.CreateWriter())
{
ds.WriteXml(writer, XmlWriteMode.WriteSchema); // XmlWriteMode.DiffGram
}
I am developing C# application, but I encountered a problem while typing into xml file. Let me show the code first:
Company comp = new Company();
comp.CompanyID = comboBox1.SelectedValue.ToString();
comp.CompanyName = comboBox1.Text;
comp.Serial = strEncryptedData;
comp.ListProduct = ll;
XmlDocument xDoc = new XmlDocument();
using (StringWriter stringWriter = new StringWriter())
{
XmlSerializer serializer = new XmlSerializer(typeof(Company));
serializer.Serialize(stringWriter, comp);
xDoc.LoadXml(stringWriter.ToString());
}
string temp = xDoc.OuterXml;
MessageBox.Show(temp);
System.IO.StreamWriter sw = new System.IO.StreamWriter(#"c:\test.xml");
sw.WriteLine(temp);
sw.Flush();
sw.Close();
Program writes the file but when i try to open it in xml format, i recieve blank document, nothing inside. When i opened it in text editor i recieve this:
<?xml version="1.0" encoding="utf-16"?><CompanyXml xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"><CompanyName /><CompanyID>100</CompanyID><Serial>00000G2SB4BER9PSFJİ2GTVM2UC1VYEİ</Serial></CompanyXml>
It is the correct data I receive, but however cannot be opened as xml.
How can I format it? Or am I doing something wrong while writing it?
just remove this text encoding="utf-16" from your first line of xml then you will open the xml.
The issue appears to be with the encoding, removing utf-16 or changing to utf-8 corrects it.
You could try the StreamWriter constructor which takes an encoding to see if it saves the .xml with the correct encoding.
For example:
StreamWriter sw = new StreamWriter(#"c:\test.xml", Encoding.UTF8);
However, i found some other way to solve this issue after you guys pointed out utf format issue.
First of all we create a class which extends to StringWriter
public class Utf8StringWriter : StringWriter
{
public override Encoding Encoding
{
get { return Encoding.UTF8; }
}
}
Then we edit code into this by modifying StringWriter:
Company comp = new Company();
comp.CompanyID = comboBox1.SelectedValue.ToString();
comp.CompanyName = comboBox1.Text;
comp.Serial = strEncryptedData;
comp.ListProduct = ll;
XmlDocument xDoc = new XmlDocument();
using (StringWriter stringWriter = new Utf8StringWriter())
{
XmlSerializer serializer = new XmlSerializer(typeof(Company));
serializer.Serialize(stringWriter, comp);
StreamWriter sw = new StreamWriter(#"c:\text.xml");
sw.WriteLine(stringWriter);
sw.Flush();
sw.Close();
}
Regards...
When I build XML up from scratch with XmlDocument, the OuterXml property already has everything nicely indented with line breaks. However, if I call LoadXml on some very "compressed" XML (no line breaks or indention) then the output of OuterXml stays that way. So ...
What is the simplest way to get beautified XML output from an instance of XmlDocument?
Based on the other answers, I looked into XmlTextWriter and came up with the following helper method:
static public string Beautify(this XmlDocument doc)
{
StringBuilder sb = new StringBuilder();
XmlWriterSettings settings = new XmlWriterSettings
{
Indent = true,
IndentChars = " ",
NewLineChars = "\r\n",
NewLineHandling = NewLineHandling.Replace
};
using (XmlWriter writer = XmlWriter.Create(sb, settings)) {
doc.Save(writer);
}
return sb.ToString();
}
It's a bit more code than I hoped for, but it works just peachy.
As adapted from Erika Ehrli's blog, this should do it:
XmlDocument doc = new XmlDocument();
doc.LoadXml("<item><name>wrench</name></item>");
// Save the document to a file and auto-indent the output.
using (XmlTextWriter writer = new XmlTextWriter("data.xml", null)) {
writer.Formatting = Formatting.Indented;
doc.Save(writer);
}
Or even easier if you have access to Linq
try
{
RequestPane.Text = System.Xml.Linq.XElement.Parse(RequestPane.Text).ToString();
}
catch (System.Xml.XmlException xex)
{
displayException("Problem with formating text in Request Pane: ", xex);
}
A shorter extension method version
public static string ToIndentedString( this XmlDocument doc )
{
var stringWriter = new StringWriter(new StringBuilder());
var xmlTextWriter = new XmlTextWriter(stringWriter) {Formatting = Formatting.Indented};
doc.Save( xmlTextWriter );
return stringWriter.ToString();
}
If the above Beautify method is being called for an XmlDocument that already contains an XmlProcessingInstruction child node the following exception is thrown:
Cannot write XML declaration.
WriteStartDocument method has already
written it.
This is my modified version of the original one to get rid of the exception:
private static string beautify(
XmlDocument doc)
{
var sb = new StringBuilder();
var settings =
new XmlWriterSettings
{
Indent = true,
IndentChars = #" ",
NewLineChars = Environment.NewLine,
NewLineHandling = NewLineHandling.Replace,
};
using (var writer = XmlWriter.Create(sb, settings))
{
if (doc.ChildNodes[0] is XmlProcessingInstruction)
{
doc.RemoveChild(doc.ChildNodes[0]);
}
doc.Save(writer);
return sb.ToString();
}
}
It works for me now, probably you would need to scan all child nodes for the XmlProcessingInstruction node, not just the first one?
Update April 2015:
Since I had another case where the encoding was wrong, I searched for how to enforce UTF-8 without BOM. I found this blog post and created a function based on it:
private static string beautify(string xml)
{
var doc = new XmlDocument();
doc.LoadXml(xml);
var settings = new XmlWriterSettings
{
Indent = true,
IndentChars = "\t",
NewLineChars = Environment.NewLine,
NewLineHandling = NewLineHandling.Replace,
Encoding = new UTF8Encoding(false)
};
using (var ms = new MemoryStream())
using (var writer = XmlWriter.Create(ms, settings))
{
doc.Save(writer);
var xmlString = Encoding.UTF8.GetString(ms.ToArray());
return xmlString;
}
}
XmlTextWriter xw = new XmlTextWriter(writer);
xw.Formatting = Formatting.Indented;
public static string FormatXml(string xml)
{
try
{
var doc = XDocument.Parse(xml);
return doc.ToString();
}
catch (Exception)
{
return xml;
}
}
A simple way is to use:
writer.WriteRaw(space_char);
Like this sample code, this code is what I used to create a tree view like structure using XMLWriter :
private void generateXML(string filename)
{
using (XmlWriter writer = XmlWriter.Create(filename))
{
writer.WriteStartDocument();
//new line
writer.WriteRaw("\n");
writer.WriteStartElement("treeitems");
//new line
writer.WriteRaw("\n");
foreach (RootItem root in roots)
{
//indent
writer.WriteRaw("\t");
writer.WriteStartElement("treeitem");
writer.WriteAttributeString("name", root.name);
writer.WriteAttributeString("uri", root.uri);
writer.WriteAttributeString("fontsize", root.fontsize);
writer.WriteAttributeString("icon", root.icon);
if (root.children.Count != 0)
{
foreach (ChildItem child in children)
{
//indent
writer.WriteRaw("\t");
writer.WriteStartElement("treeitem");
writer.WriteAttributeString("name", child.name);
writer.WriteAttributeString("uri", child.uri);
writer.WriteAttributeString("fontsize", child.fontsize);
writer.WriteAttributeString("icon", child.icon);
writer.WriteEndElement();
//new line
writer.WriteRaw("\n");
}
}
writer.WriteEndElement();
//new line
writer.WriteRaw("\n");
}
writer.WriteEndElement();
writer.WriteEndDocument();
}
}
This way you can add tab or line breaks in the way you are normally used to, i.e. \t or \n
When implementing the suggestions posted here, I had trouble with the text encoding. It seems the encoding of the XmlWriterSettings is ignored, and always overridden by the encoding of the stream. When using a StringBuilder, this is always the text encoding used internally in C#, namely UTF-16.
So here's a version which supports other encodings as well.
IMPORTANT NOTE: The formatting is completely ignored if your XMLDocument object has its preserveWhitespace property enabled when loading the document. This had me stumped for a while, so make sure not to enable that.
My final code:
public static void SaveFormattedXml(XmlDocument doc, String outputPath, Encoding encoding)
{
XmlWriterSettings settings = new XmlWriterSettings();
settings.Indent = true;
settings.IndentChars = "\t";
settings.NewLineChars = "\r\n";
settings.NewLineHandling = NewLineHandling.Replace;
using (MemoryStream memstream = new MemoryStream())
using (StreamWriter sr = new StreamWriter(memstream, encoding))
using (XmlWriter writer = XmlWriter.Create(sr, settings))
using (FileStream fileWriter = new FileStream(outputPath, FileMode.Create))
{
if (doc.ChildNodes.Count > 0 && doc.ChildNodes[0] is XmlProcessingInstruction)
doc.RemoveChild(doc.ChildNodes[0]);
// save xml to XmlWriter made on encoding-specified text writer
doc.Save(writer);
// Flush the streams (not sure if this is really needed for pure mem operations)
writer.Flush();
// Write the underlying stream of the XmlWriter to file.
fileWriter.Write(memstream.GetBuffer(), 0, (Int32)memstream.Length);
}
}
This will save the formatted xml to disk, with the given text encoding.
If you have a string of XML, rather than a doc ready for use, you can do it this way:
var xmlString = "<xml>...</xml>"; // Your original XML string that needs indenting.
xmlString = this.PrettifyXml(xmlString);
private string PrettifyXml(string xmlString)
{
var prettyXmlString = new StringBuilder();
var xmlDoc = new XmlDocument();
xmlDoc.LoadXml(xmlString);
var xmlSettings = new XmlWriterSettings()
{
Indent = true,
IndentChars = " ",
NewLineChars = "\r\n",
NewLineHandling = NewLineHandling.Replace
};
using (XmlWriter writer = XmlWriter.Create(prettyXmlString, xmlSettings))
{
xmlDoc.Save(writer);
}
return prettyXmlString.ToString();
}
A more simplified approach based on the accepted answer:
static public string Beautify(this XmlDocument doc) {
StringBuilder sb = new StringBuilder();
XmlWriterSettings settings = new XmlWriterSettings
{
Indent = true
};
using (XmlWriter writer = XmlWriter.Create(sb, settings)) {
doc.Save(writer);
}
return sb.ToString();
}
Setting the new line is not necessary. Indent characters also has the default two spaces so I preferred not to set it as well.
Set PreserveWhitespace to true before Load.
var document = new XmlDocument();
document.PreserveWhitespace = true;
document.Load(filename);