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.
We are trying to hash a xml file, i already have it working that it hashes the contents of the XML.
For which i am using the following code:
XmlDocument doc = new XmlDocument();
doc.PreserveWhitespace = true;
doc.Load(txtFile.Text);
XmlNodeList list = doc.GetElementsByTagName("Document");
XmlElement node = (XmlElement)list[0];
//node.SetAttribute("xmlns:xsi", "http://www.w3.org/2001/XMLSchema-instance");
string s = node.OuterXml;
using (MemoryStream msIn = new MemoryStream(Encoding.UTF8.GetBytes(s)))
{
XmlDsigC14NTransform t = new XmlDsigC14NTransform(true);
t.LoadInput(msIn);
using (var hash = new SHA256Managed())
{
byte[] digest = t.GetDigestedOutput(hash);
txtHash.Text = BitConverter.ToString(digest).Replace("-", String.Empty);
}
}
however, this only hashes the contents of the XML.
What i need is to hash the complete XML instead of only the contents.
If we only hash the contents, our hash doesnt compare with the control we get.
You can read the file contents without creating a XmlDocument and hash the contents:
var file = File.ReadAllBytes(txtFile.Text);
using (var hash = new SHA256Managed())
{
byte[] digest = hash.ComputeHash(file);
txtHash.Text = BitConverter.ToString(digest).Replace("-", String.Empty);
}
How to read an irregular xml string after change it to stream in C#?
Here is the code I have so far:
string modoEntrada = "01";
string xml= "";
xml = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n";
xml += "<Transaccion xmlns=\"http://www.prosa.com.mx/xsd/\">";
//xml = "<Transaccion>";
xml += "<autorizacionRequest>";
xml += "<modoEntrada>" + modoEntrada + "</modoEntrada>"; //target
xml += "</autorizacionRequest>";
xml += "</Transaccion>";
xml.Replace(" ","_");
byte[] byteArray = Encoding.UTF8.GetBytes(xml);
Stream stream = new MemoryStream(byteArray);
XmlDocument xmldoc = new XmlDocument();
xmldoc.Load(stream);
XmlNode x = xmldoc.SelectSingleNode("/autorizacionRequest/modoEntrada");
The Problem is: x == null
It works with <Transaccion> instead of <Transaccion xmlns="http://www.prosa.com.mx/xsd">.
Add a namespacemanager and use it in your xpath
XmlDocument xmldoc = new XmlDocument();
xmldoc.Load(stream);
var xmlNsM = new XmlNamespaceManager(xmldoc.NameTable);
xmlNsM.AddNamespace("tr", #"http://www.prosa.com.mx/xsd/");
XmlNode x = xmldoc.SelectSingleNode("/tr:Transaccion/tr:autorizacionRequest/tr:modoEntrada", xmlNsM );
Token is a string which I was looking for:
switch (Token)
{
case "modoEntrada":
Regex exp_01 = new Regex("<descripcion>(?<arg>[A-Z]+)?</descripcion>");
m = exp_01.Match(InputText);
break;
case "descripcionRechazo":
Regex exp_02 = new Regex("<descripcionRechazo>(?<arg>[A-Z ]+)?</descripcionRechazo>");
m = exp_02.Match(InputText);
break;
...
}
if (m.Success)
ret = m.Groups["arg"].Value;
else
ret = "";
return ret;
I have some html.
I try to clean it with library: http://sourceforge.net/projects/tidynet/
Here is my code:
//clean up html
Tidy tidy = new Tidy();
tidy.Options.DocType = DocType.Omit;
tidy.Options.DropFontTags = true;
tidy.Options.LogicalEmphasis = true;
tidy.Options.Xhtml = true;
tidy.Options.XmlOut = true;
tidy.Options.MakeClean = true;
tidy.Options.TidyMark = false;
tidy.Options.CharEncoding = CharEncoding.UTF8;
/* Declare the parameters that is needed */
TidyMessageCollection tmc = new TidyMessageCollection();
MemoryStream input = new MemoryStream();
MemoryStream output = new MemoryStream();
byte[] byteArray = Encoding.UTF8.GetBytes(report);
input.Write(byteArray, 0, byteArray.Length);
input.Position = 0;
tidy.Parse(input, output, tmc);
string cleanHtml = Encoding.UTF8.GetString(output.ToArray());
then I try to use xslt:
try
{
StringBuilder res = new StringBuilder();
XslCompiledTransform xslt = new XslCompiledTransform();
xslt.Load(XmlReader.Create(new StringReader(stylesheet.Content)));
xslt.Transform(StringExtensions.ToXmlReader(cleanHtml), null, new StringWriter(res));
var resultReport = res.ToString();
}
catch (Exception e)
{
}
and I get an exeption:
The '=' character, hexadecimal value 0x3D, cannot be included in a name
Update
How can I automaticaly clean name from '='?
HTML is not XML (unless it's XHTML). Trying to apply an XSLT to a generic HTML document is almost certainly going to give you issues. You'll have to find a different way to do whatever transformation it is that you're looking for.
If you want to process the HTML programmatically, then I would suggest the HTML Agility Pack.
I have an XML string as such:
<?xml version='1.0'?><response><error code='1'> Success</error></response>
There are no lines between one element and another, and thus is very difficult to read. I want a function that formats the above string:
<?xml version='1.0'?>
<response>
<error code='1'> Success</error>
</response>
Without resorting to manually write the format function myself, is there any .Net library or code snippet that I can use offhand?
You will have to parse the content somehow ... I find using LINQ the most easy way to do it. Again, it all depends on your exact scenario. Here's a working example using LINQ to format an input XML string.
string FormatXml(string xml)
{
try
{
XDocument doc = XDocument.Parse(xml);
return doc.ToString();
}
catch (Exception)
{
// Handle and throw if fatal exception here; don't just ignore them
return xml;
}
}
[using statements are ommitted for brevity]
Use XmlTextWriter...
public static string PrintXML(string xml)
{
string result = "";
MemoryStream mStream = new MemoryStream();
XmlTextWriter writer = new XmlTextWriter(mStream, Encoding.Unicode);
XmlDocument document = new XmlDocument();
try
{
// Load the XmlDocument with the XML.
document.LoadXml(xml);
writer.Formatting = Formatting.Indented;
// Write the XML into a formatting XmlTextWriter
document.WriteContentTo(writer);
writer.Flush();
mStream.Flush();
// Have to rewind the MemoryStream in order to read
// its contents.
mStream.Position = 0;
// Read MemoryStream contents into a StreamReader.
StreamReader sReader = new StreamReader(mStream);
// Extract the text from the StreamReader.
string formattedXml = sReader.ReadToEnd();
result = formattedXml;
}
catch (XmlException)
{
// Handle the exception
}
mStream.Close();
writer.Close();
return result;
}
This one, from kristopherjohnson is heaps better:
It doesn't require an XML document header either.
Has clearer exceptions
Adds extra behaviour options: OmitXmlDeclaration = true, NewLineOnAttributes = true
Less lines of code
static string PrettyXml(string xml)
{
var stringBuilder = new StringBuilder();
var element = XElement.Parse(xml);
var settings = new XmlWriterSettings();
settings.OmitXmlDeclaration = true;
settings.Indent = true;
settings.NewLineOnAttributes = true;
using (var xmlWriter = XmlWriter.Create(stringBuilder, settings))
{
element.Save(xmlWriter);
}
return stringBuilder.ToString();
}
The simple solution that is working for me:
XmlDocument xmlDoc = new XmlDocument();
StringWriter sw = new StringWriter();
xmlDoc.LoadXml(rawStringXML);
xmlDoc.Save(sw);
String formattedXml = sw.ToString();
Check the following link: How to pretty-print XML (Unfortunately, the link now returns 404 :()
The method in the link takes an XML string as an argument and returns a well-formed (indented) XML string.
I just copied the sample code from the link to make this answer more comprehensive and convenient.
public static String PrettyPrint(String XML)
{
String Result = "";
MemoryStream MS = new MemoryStream();
XmlTextWriter W = new XmlTextWriter(MS, Encoding.Unicode);
XmlDocument D = new XmlDocument();
try
{
// Load the XmlDocument with the XML.
D.LoadXml(XML);
W.Formatting = Formatting.Indented;
// Write the XML into a formatting XmlTextWriter
D.WriteContentTo(W);
W.Flush();
MS.Flush();
// Have to rewind the MemoryStream in order to read
// its contents.
MS.Position = 0;
// Read MemoryStream contents into a StreamReader.
StreamReader SR = new StreamReader(MS);
// Extract the text from the StreamReader.
String FormattedXML = SR.ReadToEnd();
Result = FormattedXML;
}
catch (XmlException)
{
}
MS.Close();
W.Close();
return Result;
}
I tried:
internal static void IndentedNewWSDLString(string filePath)
{
var xml = File.ReadAllText(filePath);
XDocument doc = XDocument.Parse(xml);
File.WriteAllText(filePath, doc.ToString());
}
it is working fine as expected.
.NET 2.0 ignoring name resolving, and with proper resource-disposal, indentation, preserve-whitespace and custom encoding:
public static string Beautify(System.Xml.XmlDocument doc)
{
string strRetValue = null;
System.Text.Encoding enc = System.Text.Encoding.UTF8;
// enc = new System.Text.UTF8Encoding(false);
System.Xml.XmlWriterSettings xmlWriterSettings = new System.Xml.XmlWriterSettings();
xmlWriterSettings.Encoding = enc;
xmlWriterSettings.Indent = true;
xmlWriterSettings.IndentChars = " ";
xmlWriterSettings.NewLineChars = "\r\n";
xmlWriterSettings.NewLineHandling = System.Xml.NewLineHandling.Replace;
//xmlWriterSettings.OmitXmlDeclaration = true;
xmlWriterSettings.ConformanceLevel = System.Xml.ConformanceLevel.Document;
using (System.IO.MemoryStream ms = new System.IO.MemoryStream())
{
using (System.Xml.XmlWriter writer = System.Xml.XmlWriter.Create(ms, xmlWriterSettings))
{
doc.Save(writer);
writer.Flush();
ms.Flush();
writer.Close();
} // End Using writer
ms.Position = 0;
using (System.IO.StreamReader sr = new System.IO.StreamReader(ms, enc))
{
// Extract the text from the StreamReader.
strRetValue = sr.ReadToEnd();
sr.Close();
} // End Using sr
ms.Close();
} // End Using ms
/*
System.Text.StringBuilder sb = new System.Text.StringBuilder(); // Always yields UTF-16, no matter the set encoding
using (System.Xml.XmlWriter writer = System.Xml.XmlWriter.Create(sb, settings))
{
doc.Save(writer);
writer.Close();
} // End Using writer
strRetValue = sb.ToString();
sb.Length = 0;
sb = null;
*/
xmlWriterSettings = null;
return strRetValue;
} // End Function Beautify
Usage:
System.Xml.XmlDocument xmlDoc = new System.Xml.XmlDocument();
xmlDoc.XmlResolver = null;
xmlDoc.PreserveWhitespace = true;
xmlDoc.Load("C:\Test.svg");
string SVG = Beautify(xmlDoc);
Customizable Pretty XML output with UTF-8 XML declaration
The following class definition gives a simple method to convert an input XML string into formatted output XML with the xml declaration as UTF-8. It supports all the configuration options that the XmlWriterSettings class offers.
using System;
using System.Text;
using System.Xml;
using System.IO;
namespace CJBS.Demo
{
/// <summary>
/// Supports formatting for XML in a format that is easily human-readable.
/// </summary>
public static class PrettyXmlFormatter
{
/// <summary>
/// Generates formatted UTF-8 XML for the content in the <paramref name="doc"/>
/// </summary>
/// <param name="doc">XmlDocument for which content will be returned as a formatted string</param>
/// <returns>Formatted (indented) XML string</returns>
public static string GetPrettyXml(XmlDocument doc)
{
// Configure how XML is to be formatted
XmlWriterSettings settings = new XmlWriterSettings
{
Indent = true
, IndentChars = " "
, NewLineChars = System.Environment.NewLine
, NewLineHandling = NewLineHandling.Replace
//,NewLineOnAttributes = true
//,OmitXmlDeclaration = false
};
// Use wrapper class that supports UTF-8 encoding
StringWriterWithEncoding sw = new StringWriterWithEncoding(Encoding.UTF8);
// Output formatted XML to StringWriter
using (XmlWriter writer = XmlWriter.Create(sw, settings))
{
doc.Save(writer);
}
// Get formatted text from writer
return sw.ToString();
}
/// <summary>
/// Wrapper class around <see cref="StringWriter"/> that supports encoding.
/// Attribution: http://stackoverflow.com/a/427737/3063884
/// </summary>
private sealed class StringWriterWithEncoding : StringWriter
{
private readonly Encoding encoding;
/// <summary>
/// Creates a new <see cref="PrettyXmlFormatter"/> with the specified encoding
/// </summary>
/// <param name="encoding"></param>
public StringWriterWithEncoding(Encoding encoding)
{
this.encoding = encoding;
}
/// <summary>
/// Encoding to use when dealing with text
/// </summary>
public override Encoding Encoding
{
get { return encoding; }
}
}
}
}
Possibilities for further improvement:-
An additional method GetPrettyXml(XmlDocument doc, XmlWriterSettings settings) could be created that allows the caller to customize the output.
An additional method GetPrettyXml(String rawXml) could be added that supports parsing raw text, rather than have the client use the XmlDocument. In my case, I needed to manipulate the XML using the XmlDocument, hence I didn't add this.
Usage:
String myFormattedXml = null;
XmlDocument doc = new XmlDocument();
try
{
doc.LoadXml(myRawXmlString);
myFormattedXml = PrettyXmlFormatter.GetPrettyXml(doc);
}
catch(XmlException ex)
{
// Failed to parse XML -- use original XML as formatted XML
myFormattedXml = myRawXmlString;
}
Check the following link: Format an XML file so it looks nice in C#
// Format the XML text.
StringWriter string_writer = new StringWriter();
XmlTextWriter xml_text_writer = new XmlTextWriter(string_writer);
xml_text_writer.Formatting = Formatting.Indented;
xml_document.WriteTo(xml_text_writer);
// Display the result.
txtResult.Text = string_writer.ToString();
It is possible to pretty-print an XML string via a streaming transformation with XmlWriter.WriteNode(XmlReader, true). This method
copies everything from the reader to the writer and moves the reader to the start of the next sibling.
Define the following extension methods:
public static class XmlExtensions
{
public static string FormatXml(this string xml, bool indent = true, bool newLineOnAttributes = false, string indentChars = " ", ConformanceLevel conformanceLevel = ConformanceLevel.Document) =>
xml.FormatXml( new XmlWriterSettings { Indent = indent, NewLineOnAttributes = newLineOnAttributes, IndentChars = indentChars, ConformanceLevel = conformanceLevel });
public static string FormatXml(this string xml, XmlWriterSettings settings)
{
using (var textReader = new StringReader(xml))
using (var xmlReader = XmlReader.Create(textReader, new XmlReaderSettings { ConformanceLevel = settings.ConformanceLevel } ))
using (var textWriter = new StringWriter())
{
using (var xmlWriter = XmlWriter.Create(textWriter, settings))
xmlWriter.WriteNode(xmlReader, true);
return textWriter.ToString();
}
}
}
And now you will be able to do:
var inXml = #"<?xml version='1.0'?><response><error code='1'> Success</error></response>";
var newXml = inXml.FormatXml(indentChars : "", newLineOnAttributes : false); // Or true, if you prefer
Console.WriteLine(newXml);
Which prints
<?xml version='1.0'?>
<response>
<error code="1"> Success</error>
</response>
Notes:
Other answers load the XML into some Document Object Model such as XmlDocument or XDocument/XElement, then re-serialize the DOM with indentation enabled.
This streaming solution completely avoids the added memory overhead of a DOM.
In your question you do not add any indentation for the nested <error code='1'> Success</error> node, so I set indentChars : "". Generally an indentation of two spaces per level of nesting is customary.
Attribute delimiters will be unconditionally transformed to double-quotes if currently single-quotes. (I believe this is true of other answers as well.)
Passing conformanceLevel : ConformanceLevel.Fragment allows strings containing sequences of XML fragments to be formatted.
Other than ConformanceLevel.Fragment, the input XML string must be well-formed. If it is not, XmlReader will throw an exception.
Demo fiddle here.
if you load up the XMLDoc I'm pretty sure the .ToString() function posses an overload for this.
But is this for debugging? The reason that it is sent like that is to take up less space (i.e stripping unneccessary whitespace from the XML).
Hi why don't you just try this:
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.PreserveWhitespace = false;
....
....
xmlDoc.Save(fileName);
PreserveWhitespace = false; that option can be used xml beautifier as well.