I want to deserialize an xml with this code:
XmlSerializer serializer = new XmlSerializer(typeof(TrafficJunctions));
using (StringReader reader = new StringReader(XmlPath))
{
var test = (TrafficJunctions)serializer.Deserialize(reader);
}
But get the following exception:
XmlException: Data at the root level is invalid. Line 1, position 1.
Apparently there is a BOM at the beginning of the xml. I know that I have to remove it, but I don't understand how to do it for before deserialization. Any help is welcome.
Thanks in advance.
Solution (Credit to canton7):
XmlSerializer serializer = new XmlSerializer(typeof(TrafficJunctions));
using (StreamReader reader = new StreamReader(XmlPath))
{
var test = (TrafficJunctions)serializer.Deserialize(reader);
}
Related
I'm currently trying to serialize a class into XML to be posted to php web service.
Whenever I did the normal serialization using XMLSerializer, XML declaration is always appear in the first line of the XML document (similar as to <?xml ....?>). I tested the XML and unable to get it working because the endpoint does not accept XML declaration and I can't do anything about it.
I'm unfamiliar with XML Serialization in C# to be honest.
Therefore, I used XMLWriter to do this as below :-
private string SerializeClassToString(GetRiskReport value)
{
var emptyNS = new XmlSerializerNamespaces(new[] { XmlQualifiedName.Empty });
var ser = new XmlSerializer(value.GetType());
var settings = new XmlWriterSettings();
settings.OmitXmlDeclaration = true;
using (var stream = new StringWriter())
{
using (var writer = XmlWriter.Create(stream, settings))
{
ser.Serialize(writer, value, emptyNS);
return stream.ToString();
}
}
}
Result for the Namespace is
<GetRiskReport FCRA=\"false\" ReturnResultsOnly=\"false\" Monitoring=\"false\">
... and I'm able to omit the XML Declaration, however I'm being introduced with 2 new problem.
I got \r\n for new line and I have escaped double quote such as ReturnResultsOnly=\"false\" Monitoring=\"false\" which is also unable processed by the endpoint.
I would like to ask is that does anyone can give me an idea on how to change the XmlWriterSetting to omit XML Declaration, avoid \r\n and also avoid escaped double quotes \"
Thanks for your advice in advance.
Simon
Try with following settings
settings.NewLineHandling = NewLineHandling.None;
settings.CheckCharacters = false;
private void SerializeClassToString(GetRiskReport value)
{
var emptyNS = new XmlSerializerNamespaces(new[]{XmlQualifiedName.Empty});
var ser = new XmlSerializer(value.GetType());
var settings = new XmlWriterSettings();
settings.OmitXmlDeclaration = true;
string path = 'your_file_path_here'
if (File.Exists(path)) File.Delete(path);
FileStream stream = File.Create(path);
using (var writer = XmlWriter.Create(stream, settings))
{
ser.Serialize(writer, value, emptyNS);
return;
}
}
There was no way to avoid ms bug or thier intensional specification about xmlserializing.It's easier and faster to use filestream object.
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 am new for Json and I have a simple problem.
I am trying to convert json file to xml file with c#. But it throw an exception.
The Code is ;
private void TakeXML()
{
string json = ReadText();
XmlDocument doc = (XmlDocument)Newtonsoft.Json.JsonConvert.DeserializeXmlNode(json);
XmlTextWriter writer = new XmlTextWriter("json.xml", null);
writer.Formatting = Formatting.Indented;
doc.Save(writer);
}
The ReadText function is;
private string ReadText()
{
FileStream fs = new FileStream(#"C:\Users\Sinan\Desktop\bina.json", FileMode.Open, FileAccess.Read);
StreamReader sr = new StreamReader(fs);
string json;
try
{
json = sr.ReadToEnd();
return json;
}
catch (Exception)
{
return null;
}
finally
{
sr.Close();
fs.Dispose();
}
}
for XmlDocument doc = (XmlDocument)Newtonsoft.Json.JsonConvert.DeserializeXmlNode(json); line, it said that;
"JSON root object has multiple properties. The root object must have a single property in order to create a valid XML document. Consider specifing a DeserializeRootElementName."
I am searching to solve this problem but ı haven't found it. İf you help me in this regard, I will be glad. Thank you.
In method DeserializeXmlNode specify the root node name in second parameter as shown in below code:
XmlDocument doc =
(XmlDocument)
Newtonsoft.Json.JsonConvert.DeserializeXmlNode(json, "rootNodeName");
// second parameter
Although if you can give json string then it would be easy to give exact answer.
Reference link: Converting JSON to XML
I have been working with XML in database LINQ and find that it is very difficult to work with the serializer.
The database LINQ required a field that store XElement.
I have a complex object with many customized structure class, so I would like to use the XmlSerializer to serialize the object.
However, the serializer can only serialize to file ("C:\xxx\xxx.xml") or a memory stream.
However to convert or serialize it to be a XElement so that I can store in the database using LINQ?
And How to do the reverse? i.e. Deserialize an XElement...
Try to use this
using (var stream = new MemoryStream())
{
serializer.Serialize(stream, value);
stream.Position = 0;
using (XmlReader reader = XmlReader.Create(stream))
{
XElement element = XElement.Load(reader);
}
}
deserialize :
XmlSerializer xs = new XmlSerializer(typeof(XElement));
using (MemoryStream ms = new MemoryStream())
{
xs.Serialize(ms, xml);
ms.Position = 0;
xs = new XmlSerializer(typeof(YourType));
object obj = xs.Deserialize(ms);
}
To make what John Saunders was describing more explicit, deserialization is very straightforward:
public static object DeserializeFromXElement(XElement element, Type t)
{
using (XmlReader reader = element.CreateReader())
{
XmlSerializer serializer = new XmlSerializer(t);
return serializer.Deserialize(reader);
}
}
Serialization is a little messier because calling CreateWriter() from an XElement or XDocument creates child elements. (In addition, the XmlWriter created from an XElement has ConformanceLevel.Fragment, which causes XmlSerialize to fail unless you use the workaround here.) As a result, I use an XDocument, since this requires a single element, and gets us around the XmlWriter issue:
public static XElement SerializeToXElement(object o)
{
var doc = new XDocument();
using (XmlWriter writer = doc.CreateWriter())
{
XmlSerializer serializer = new XmlSerializer(o.GetType());
serializer.Serialize(writer, o);
}
return doc.Root;
}
First of all, see Serialize Method to see that the serializer can handle alot more than just memory streams or files.
Second, try using XElement.CreateWriter and then passing the resulting XmlWriter to the serializer.
The SQL has XML data type may be this can help you look at msdn
I have the following extension method to serialize my class....
public static string SerializeToXml<T>(this object obj)
{
XDocument doc = new XDocument();
XmlSerializer ser = new XmlSerializer(typeof(T));
using (var writer = doc.CreateWriter())
{
ser.Serialize(writer, obj);
}
return doc.ToString();
}
This seems to work fine and returns the following string for my serialized object:
<AuthenticatedUser xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Username>mark</Username>
<UserID>4</UserID>
<Roles>
<string>AuthenticatedUsers</string>
</Roles>
<IsValid>false</IsValid>
</AuthenticatedUser>
However when I try to deserialize this string using the method below I get this error:
{"The encoding style '<AuthenticatedUser xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\">\r\n <Username>mark</Username>\r\n <UserID>4</UserID>\r\n <Roles>\r\n <string>AuthenticatedUsers</string>\r\n </Roles>\r\n <IsMale>false</IsMale>\r\n</AuthenticatedUser>' is not valid for this call because this XmlSerializer instance does not support encoding. Use the SoapReflectionImporter to initialize an XmlSerializer that supports encoding."}
Method....
public static T DeserializeFromXml<T>(this string xml)
{
var element = XElement.Parse(xml);
XmlSerializer ser = new XmlSerializer(typeof(T));
using (var reader = element.CreateReader())
{
return (T)ser.Deserialize(reader, xml);
}
}
So after I read the error message I changed the deserialize method to use the SoadReflectionImporter....
public static T DeserializeFromXml<T>(this string xml)
{
var element = XElement.Parse(xml);
SoapReflectionImporter soap = new SoapReflectionImporter();
var mapping = soap.ImportTypeMapping(typeof(T));
XmlSerializer ser = new XmlSerializer(mapping);
using (var reader = element.CreateReader())
{
return (T)ser.Deserialize(reader, xml);
}
}
However I now get this error...
{"The encoding style '<AuthenticatedUser xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\">\r\n <Username>mark</Username>\r\n <UserID>4</UserID>\r\n <Roles>\r\n <string>AuthenticatedUsers</string>\r\n </Roles>\r\n <IsValid>false</IsValid>\r\n</AuthenticatedUser>' is not valid for this call. Valid values are 'http://schemas.xmlsoap.org/soap/encoding/' for SOAP 1.1 encoding or 'http://www.w3.org/2003/05/soap-encoding' for SOAP 1.2 encoding."}
Does anyone know where I'm going wrong and how I can deserialize this string successfully?
The problem is the overload of the Deserialize method that you are calling:
return (T)ser.Deserialize(reader, xml);
The xml parameter in the call specifies the encoding style, but in this case you are passing the xml from the serialization. Simply delete the second parameter and just call Deserialize with the reader and it should work fine:
return (T)ser.Deserialize(reader);
XElement.CreateReader() doesn't return the XDeclaration.
Instead, try making an XmlReader from a StringReader.
Why are you using XmlSerializer ?
Unless you must control the way the output XML looks, you should be using DataContractSerializer
Here is a nice blog post about the two
Do you need the Parse(xml) call and the reader element? Since you have the string, can't you just deserialize the string? First convert to bytes...
byte [] bytes = Encoding.Unicode.GetBytes(xml);
MemoryStream mem = new MemoryStream(bytes);
returnValue = (T)ser.Deserialize(mem);