I have created the following wrapper method to disable DTD
public class Program
{
public static void Main(string[] args)
{
string s = #"<?xml version =""1.0"" encoding=""utf-16""?>
<ArrayOfSerializingTemplateItem xmlns:xsd=""http://www.w3.org/2001/XMLSchema"" xmlns:xsi=""http://www.w3.org/2001/XMLSchema-instance"">
<SerializingTemplateItem>
</SerializingTemplateItem>
</ArrayOfSerializingTemplateItem >";
try
{
XmlReader reader = XmlWrapper.CreateXmlReaderObject(s);
XmlSerializer sr = new XmlSerializer(typeof(List<SerializingTemplateItem>));
Object ob = sr.Deserialize(reader);
}
catch (Exception ex)
{
Console.WriteLine(ex);
throw;
}
Console.ReadLine();
}
}
public class XmlWrapper
{
public static XmlReader CreateXmlReaderObject(string sr)
{
byte[] byteArray = Encoding.UTF8.GetBytes(sr);
MemoryStream stream = new MemoryStream(byteArray);
stream.Position = 0;
XmlReaderSettings settings = new XmlReaderSettings();
settings.ValidationType = ValidationType.None;
settings.DtdProcessing = DtdProcessing.Ignore;
return XmlReader.Create(stream, settings);
}
}
public class SerializingTemplateItem
{
}
The above throws exception "There is no Unicode byte order mark. Cannot switch to Unicode." (Demo fiddle here: https://dotnetfiddle.net/pGxOE9).
But if I use the following code to create the XmlReader instead of calling the XmlWrapper method. It works fine.
StringReader stringReader = new StringReader( xml );
XmlReader reader = new XmlTextReader( stringReader );
But I need to use the wrapper method as a security requirement to disable DTD. I don't know why I am unable to deserialize after calling my wrapper method. Any help will be highly appreciated.
Your problem is that you have encoded the XML into a MemoryStream using Encoding.UTF8, but the XML string itself claims to be encoded in UTF-16 in the encoding declaration in its XML text declaration:
<?xml version ="1.0" encoding="utf-16"?>
<ArrayOfSerializingTemplateItem>
<!-- Content omitted -->
</ArrayOfSerializingTemplateItem >
Apparently when the XmlReader encounters this declaration, it tries honor the declaration and switch from UTF-8 to UTF-16 but fails for some reason - possibly because the stream really is encoded in UTF-8. Conversely when the deprecated XmlTextReader encounters the declaration, it apparently just ignores it as not implemented, which happens to cause things to work successfully in this situation.
The simplest way to resolve this is to read directly from the string using a StringReader using XmlReader.Create(TextReader, XmlReaderSettings):
public class XmlWrapper
{
public static XmlReader CreateXmlReaderObject(string sr)
{
var settings = new XmlReaderSettings
{
ValidationType = ValidationType.None,
DtdProcessing = DtdProcessing.Ignore,
};
return XmlReader.Create(new StringReader(sr), settings);
}
}
Since a c# string is always encoded internally in UTF-16 the encoding statement in the XML will be ignored as irrelevant. This will also be more performant as the conversion to an intermediate byte array is completely skipped.
Incidentally, you should dispose of your XmlReader via a using statement:
Object ob;
using (var reader = XmlWrapper.CreateXmlReaderObject(s))
{
XmlSerializer sr = new XmlSerializer(typeof(List<SerializingTemplateItem>));
ob = sr.Deserialize(reader);
}
Working sample fiddle here.
Related questions:
Meaning of - <?xml version="1.0" encoding="utf-8"?>
Ignoring specified encoding when deserializing XML
I have a problem when I deserialize the xml into List of Objects. I searched it on the net this morning, but my problem isn't resolved.
Deserialization method
public static List<FileAction> DeSerialization()
{
XmlRootAttribute xRoot=new XmlRootAttribute();
xRoot.ElementName="ArrayOfSerializeClass";
xRoot.IsNullable=true;
XmlSerializer serializer = new XmlSerializer(typeof(List<FileAction>),xRoot);//, new XmlRootAttribute("ArrayOfSerializeClass")
using (Stream streamReader = File.OpenRead(#"C:\serialization\SerializationWithFileWatcher\Output\XmlSerialize.xml"))//FileStream fs =new FileStream(xmlPath,FileMode.Open)
{
using (XmlReader reader = XmlReader.Create(streamReader))
{
int count =0;
List<FileAction> serialList2 = (List<FileAction>)serializer.Deserialize(reader);
return (List<FileAction>)serializer.Deserialize(reader);
}
}
Calling Method
String resultPath = #"C:\serialization\SerializationWithFileWatcher\Output\XmlSerialize.xml";
if (!File.Exists(resultPath))
{
XmlSerializer xs = new XmlSerializer(typeof(List<SerializeClass>));
using (FileStream fileStream = new FileStream(#"C:\serialization\SerializationWithFileWatcher\Output\XmlSerialize.xml", FileMode.Create))
{
xs.Serialize(fileStream, serializeList);//seri
fileStream.Close();
}
Console.WriteLine("Succesfully serialized to XML");
}
else
{
//string path= #"C:\serialization\SerializationWithFileWatcher\Output\XmlSerialize.xml";
DeSerialization();
XmlSerializer xs = new XmlSerializer(typeof(List<SerializeClass>));
FileStream fs = new FileStream(#"C:\serialization\SerializationWithFileWatcher\Output\XmlSerialize.xml", FileMode.Append, FileAccess.Write);
using (XmlWriter xwr = XmlWriter.Create(fs))//TextWriter xwr = new StreamWriter
{
xs.Serialize(xwr, serializeList);//seri
//fs.Close();
}
Console.WriteLine("Succesfully serialized to XML");
}
return serializeList;
The reason why I am calling it here is that I want to add this object again to the xml file.
THe error is that here is an error in XML document (15,27).
My Xml structure
<?xml version="1.0"?>
<ArrayOfSerializeClass xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<SerializeClass>
<creationTime>2013-11-25T09:53:25.3325289+05:30</creationTime>
<fileAction>Renamed</fileAction>
<Properties>
<FileAttributes fileName="validate json.txt">
<fileSize>307</fileSize>
<extension>.txt</extension>
<lastAccessTime>2013-11-25T09:53:25.3325289+05:30</lastAccessTime
<fullPath>C:\serialization\SerializationWithFileWatcher\SerializationWithFileWatcherProj\validate json.txt</fullPath>
</FileAttributes>
</Properties>
</SerializeClass>
</ArrayOfSerializeClass>
What I understand from the code above is that you are trying to extend the current XML, by first reading it as a FileStream and then using an XmlWriter to add some more content to it.
If my understanding is correct, then you are trying to write to the end of an existing XML file, which is not allowed since any XML document can have only one root node. In your case, that root node is ArrayOfSerializeClass.
So, in order to successfully achieve your task, you must append your XML within the root node.
Update:
Possible solution here: how to append a xml file in c#?
I have an xml file that is styled with xslt to produce html email template. I get values from users dynamically and replace xml elements text with the values received. How can i get that xml file and send as html email body in c#?
My Xml Looks Like This
<?xml version="1.0" encoding="utf-8" ?>
<?xml-stylesheet type="text/xsl" href="EmailTemplateStyleForHTML.xslt"?>
<EmailTemplate>
<subject>Information from xyz</subject>
<displayName>abcd</displayName>
<Message1>
Thanks you for registering to xyz.
</Message1>
<Copyright>Copyright xyz</Copyright>
</EmailTemplate>
I am using LINQ to set the values to the xml.
Note: I know how to get and set the values from xml but what i want is to grab the whole xml file in to the email body.
If you think there is a better approach for this i would love to hear that. I would really appreciate your help.
Edited after Reply:
The xsl transformation part :
TextReader tr1 = new StringReader(#"EMailTemplateHtml.xml");
var tr11 = new XmlTextReader(tr1);
var xPathDocument = new XPathDocument(tr11);
//read XSLT
TextReader tr2 = new StringReader(#"EmailTemplateStyleForHTML.xslt");
var tr22 = new XmlTextReader(tr2);
var xslt = new XslTransform();
xslt.Load(tr22);
var sb = new StringBuilder();
TextWriter tw = new StringWriter(sb);
xslt.Transform(xPathDocument, null, tw);
emailBody = sb.ToString();
I am doing the transformation as you said(#Roy Ashbrook) am i missing anything here?
I believe you will need to actually perform the XSL transform in memory, not reference it in the XML itself. It's possible you could store the XSL in a remote location and reference it that way, but I wouldn't
so:
inject your values into your xml string
transform your xml using your xsl
make that your html message body
Here is some code. Mostly borrowed from/inspired by:
How to transform XML as a string w/o using files in .NET? and Sending E-mail using C#.
void Main()
{
SendHtmlBody(GetHtmlBody());
}
void SendHtmlBody(string HtmlBody){
using(SmtpClient c = new SmtpClient())
{
//set smtp options here
using(MailMessage msg = new MailMessage("from#replace.me","to#replace.me"))
{
msg.Subject = "Testing Bulk mail application";
msg.Body = HtmlBody;
msg.IsBodyHtml = true;
//c.Send(msg);
}
}
}
string GetHtmlBody(){
string xmlInput = #"<?xml version=""1.0"" encoding=""utf-8"" ?>
<EmailTemplate>
<subject>Information from xyz</subject>
<displayName>abcd</displayName>
<Message1>
Thanks you for registering to xyz.
</Message1>
<Copyright>Copyright xyz</Copyright>
</EmailTemplate>";
string xslInput = #"<?xml version=""1.0"" encoding=""ISO-8859-1""?>
<xsl:stylesheet version=""1.0""
xmlns:xsl=""http://www.w3.org/1999/XSL/Transform"">
<xsl:template match=""/"">
<html>
<body>
<h5><xsl:value-of select=""EmailTemplate/subject""/></h5>
<h5><xsl:value-of select=""EmailTemplate/displayName""/></h5>
</body>
</html>
</xsl:template>
</xsl:stylesheet>";
using (StringReader srt = new StringReader(xslInput)) // xslInput is a string that contains xsl
using (StringReader sri = new StringReader(xmlInput)) // xmlInput is a string that contains xml
{
using (XmlReader xrt = XmlReader.Create(srt))
using (XmlReader xri = XmlReader.Create(sri))
{
XslCompiledTransform xslt = new XslCompiledTransform();
xslt.Load(xrt);
using (StringWriter sw = new StringWriter())
using (XmlWriter xwo = XmlWriter.Create(sw, xslt.OutputSettings)) // use OutputSettings of xsl, so it can be output as HTML
{
xslt.Transform(xri, xwo);
return sw.ToString();
}
}
}
}
I have the following code:
string body = "<custom xml>";
XDocument doc = XDocument.Parse(body);
MemoryStream stream = new MemoryStream();
XmlWriter writer = XmlWriter.Create(stream);
if (writer != null)
{
doc.Save(writer);
writer.Flush();
writer.Close();
}
stream.Position = 0;
XmlReader rd = XmlReader.Create(stream);
Message output = Message.CreateMessage(msg.Version, msg.Headers.Action, rd);
output.Headers.CopyHeadersFrom(msg);
output.Properties.CopyProperties(msg.Properties);
When I try to use the message I get the following error:
hexadecimal value 0x02, is an invalid character. Line 1, position 2.
Any idea why? And what I can do to fix this?
Try something like this:
string body = "<?xml version='1.0'?><custom></custom>";
First of all, you often need the <?xml version='1.0'?> header, and as Marc G. already mentioned, your <custom xml> is not valid XML; first of all, XML tags cannot contain spaces, and second of all the opened tag is never closed.
How can I generate valid XML in C#?
It depends on the scenario. XmlSerializer is certainly one way and has the advantage of mapping directly to an object model. In .NET 3.5, XDocument, etc. are also very friendly. If the size is very large, then XmlWriter is your friend.
For an XDocument example:
Console.WriteLine(
new XElement("Foo",
new XAttribute("Bar", "some & value"),
new XElement("Nested", "data")));
Or the same with XmlDocument:
XmlDocument doc = new XmlDocument();
XmlElement el = (XmlElement)doc.AppendChild(doc.CreateElement("Foo"));
el.SetAttribute("Bar", "some & value");
el.AppendChild(doc.CreateElement("Nested")).InnerText = "data";
Console.WriteLine(doc.OuterXml);
If you are writing a large stream of data, then any of the DOM approaches (such as XmlDocument/XDocument, etc.) will quickly take a lot of memory. So if you are writing a 100 MB XML file from CSV, you might consider XmlWriter; this is more primitive (a write-once firehose), but very efficient (imagine a big loop here):
XmlWriter writer = XmlWriter.Create(Console.Out);
writer.WriteStartElement("Foo");
writer.WriteAttributeString("Bar", "Some & value");
writer.WriteElementString("Nested", "data");
writer.WriteEndElement();
Finally, via XmlSerializer:
[Serializable]
public class Foo
{
[XmlAttribute]
public string Bar { get; set; }
public string Nested { get; set; }
}
...
Foo foo = new Foo
{
Bar = "some & value",
Nested = "data"
};
new XmlSerializer(typeof(Foo)).Serialize(Console.Out, foo);
This is a nice model for mapping to classes, etc.; however, it might be overkill if you are doing something simple (or if the desired XML doesn't really have a direct correlation to the object model). Another issue with XmlSerializer is that it doesn't like to serialize immutable types : everything must have a public getter and setter (unless you do it all yourself by implementing IXmlSerializable, in which case you haven't gained much by using XmlSerializer).
The best thing hands down that I have tried is LINQ to XSD (which is unknown to most developers). You give it an XSD Schema and it generates a perfectly mapped complete strongly-typed object model (based on LINQ to XML) for you in the background, which is really easy to work with - and it updates and validates your object model and XML in real-time. While it's still "Preview", I have not encountered any bugs with it.
If you have an XSD Schema that looks like this:
<xs:element name="RootElement">
<xs:complexType>
<xs:sequence>
<xs:element name="Element1" type="xs:string" />
<xs:element name="Element2" type="xs:string" />
</xs:sequence>
<xs:attribute name="Attribute1" type="xs:integer" use="optional" />
<xs:attribute name="Attribute2" type="xs:boolean" use="required" />
</xs:complexType>
</xs:element>
Then you can simply build XML like this:
RootElement rootElement = new RootElement;
rootElement.Element1 = "Element1";
rootElement.Element2 = "Element2";
rootElement.Attribute1 = 5;
rootElement.Attribute2 = true;
Or simply load an XML from file like this:
RootElement rootElement = RootElement.Load(filePath);
Or save it like this:
rootElement.Save(string);
rootElement.Save(textWriter);
rootElement.Save(xmlWriter);
rootElement.Untyped also yields the element in form of a XElement (from LINQ to XML).
new XElement("Foo",
from s in nameValuePairList
select
new XElement("Bar",
new XAttribute("SomeAttr", "SomeAttrValue"),
new XElement("Name", s.Name),
new XElement("Value", s.Value)
)
);
XmlWriter is the fastest way to write good XML. XDocument, XMLDocument and some others works good aswell, but are not optimized for writing XML. If you want to write the XML as fast as possible, you should definitely use XmlWriter.
In the past I have created my XML Schema, then used a tool to generate C# classes which will serialize to that schema. The XML Schema Definition Tool is one example
http://msdn.microsoft.com/en-us/library/x6c1kb0s(VS.71).aspx
I think this resource should suffice for a moderate XML save/load: Read/Write XML using C#.
My task was to store musical notation. I choose XML, because I guess .NET has matured enough to allow easy solution for the task. I was right :)
This is my song file prototype:
<music judul="Kupu-Kupu yang Lucu" pengarang="Ibu Sud" tempo="120" birama="4/4" nadadasar="1=F" biramapembilang="4" biramapenyebut="4">
<not angka="1" oktaf="0" naikturun="" nilai="1"/>
<not angka="2" oktaf="0" naikturun="" nilai="0.5"/>
<not angka="5" oktaf="1" naikturun="/" nilai="0.25"/>
<not angka="2" oktaf="0" naikturun="\" nilai="0.125"/>
<not angka="1" oktaf="0" naikturun="" nilai="0.0625"/>
</music>
That can be solved quite easily:
For Save to File:
private void saveToolStripMenuItem_Click(object sender, EventArgs e)
{
saveFileDialog1.Title = "Save Song File";
saveFileDialog1.Filter = "Song Files|*.xsong";
if (saveFileDialog1.ShowDialog() == DialogResult.OK)
{
FileStream fs = new FileStream(saveFileDialog1.FileName, FileMode.Create);
XmlTextWriter w = new XmlTextWriter(fs, Encoding.UTF8);
w.WriteStartDocument();
w.WriteStartElement("music");
w.WriteAttributeString("judul", Program.music.getTitle());
w.WriteAttributeString("pengarang", Program.music.getAuthor());
w.WriteAttributeString("tempo", Program.music.getTempo()+"");
w.WriteAttributeString("birama", Program.music.getBirama());
w.WriteAttributeString("nadadasar", Program.music.getNadaDasar());
w.WriteAttributeString("biramapembilang", Program.music.getBiramaPembilang()+"");
w.WriteAttributeString("biramapenyebut", Program.music.getBiramaPenyebut()+"");
for (int i = 0; i < listNotasi.Count; i++)
{
CNot not = listNotasi[i];
w.WriteStartElement("not");
w.WriteAttributeString("angka", not.getNot() + "");
w.WriteAttributeString("oktaf", not.getOktaf() + "");
String naikturun="";
if(not.isTurunSetengah())naikturun="\\";
else if(not.isNaikSetengah())naikturun="/";
w.WriteAttributeString("naikturun",naikturun);
w.WriteAttributeString("nilai", not.getNilaiNot()+"");
w.WriteEndElement();
}
w.WriteEndElement();
w.Flush();
fs.Close();
}
}
For load file:
openFileDialog1.Title = "Open Song File";
openFileDialog1.Filter = "Song Files|*.xsong";
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
FileStream fs = new FileStream(openFileDialog1.FileName, FileMode.Open);
XmlTextReader r = new XmlTextReader(fs);
while (r.Read())
{
if (r.NodeType == XmlNodeType.Element)
{
if (r.Name.ToLower().Equals("music"))
{
Program.music = new CMusic(r.GetAttribute("judul"),
r.GetAttribute("pengarang"),
r.GetAttribute("birama"),
Convert.ToInt32(r.GetAttribute("tempo")),
r.GetAttribute("nadadasar"),
Convert.ToInt32(r.GetAttribute("biramapembilang")),
Convert.ToInt32(r.GetAttribute("biramapenyebut")));
}
else
if (r.Name.ToLower().Equals("not"))
{
CNot not = new CNot(Convert.ToInt32(r.GetAttribute("angka")), Convert.ToInt32(r.GetAttribute("oktaf")));
if (r.GetAttribute("naikturun").Equals("/"))
{
not.setNaikSetengah();
}
else if (r.GetAttribute("naikturun").Equals("\\"))
{
not.setTurunSetengah();
}
not.setNilaiNot(Convert.ToSingle(r.GetAttribute("nilai")));
listNotasi.Add(not);
}
}
else
if (r.NodeType == XmlNodeType.Text)
{
Console.WriteLine("\tVALUE: " + r.Value);
}
}
}
}
}
For simple things, I just use the XmlDocument/XmlNode/XmlAttribute classes and XmlDocument DOM found in System.XML.
It generates the XML for me, I just need to link a few items together.
However, on larger things, I use XML serialization.
For simple cases, I would also suggest looking at XmlOutput a fluent interface for building Xml.
XmlOutput is great for simple Xml creation with readable and maintainable code, while generating valid Xml. The orginal post has some great examples.
As above.
I use stringbuilder.append().
Very straightforward, and you can then do xmldocument.load(strinbuilder object as parameter).
You will probably find yourself using string.concat within the append parameter, but this is a very straightforward approach.