I am trying to generate an XML file from an Object. I put a break point before generating the XML file, so I can check the values. The object and its value look fine. However, after the XML file is generated, it is missing a key component, the code.
This is what I was expected to see.
<eDocument Code="UN" Cat="ST">
<id myId="5"/>
</eDocument>
This is the actual xml file that is generated.
<eDocument Cat="EST">
<id myId="5"/>
</eDocument>
This is the object that is being serialize to generate the xml file.
sDoc eDocument = new sDoc();
eDocument.Code = "UN";
eDocument.Cat = "ST";
eDocument.myId = new ID[1];
eDocument.myId[0].id= 5;
This is how I am saving the file
string fileName= "student.xml";
XmlSerializer serializeObject = new XmlSerializer(eDocument.GetType());
TextWriter streamWritter = new StreamWriter(Server.MapPath(#"~/student/" + fileName));
serializeObject.Serialize(streamWritter, eDocument); // I check the eDocument Object, and it has all the correct inforamtion
streamWritter.Close();
Is there something that I am doing wrong here?
You need to check the "Code" property in the sDoc class.
A property should be public read/write in order to be Xml-serializable. By default, if no attribute is applied to a public property, it is serialized as an XML element. In your case it's not serialized at all, which means something is wrong.
First check: is the property public in both read (get) and write (set)?
Second check: Isn't the field marked with [XmlIgnoreAttribute]?
And finally: Marking the Cat property with [XmlAttribute] will xml-serialize it as an attribute.
Related
I have an object that I serialize into an XML file. Everything works as expected, but I'm wanting to add a Version attribute to the root element.
What's the best way to do this?
Here's an example of how I'm serializing:
MyProgram newProgram = new MyProgram()
{
ValueA = "A value.",
ValueB = "B value.",
ValueC = "C value."
};
XmlSerializer xmlSerializer = new XmlSerializer(typeof(MyProgram));
StreamWriter streamWriter = new StreamWriter(fileName);
xmlSerializer.Serialize(streamWriter, newProgram);
streamWriter.Close();
Right now, my XML looks something like this:
<MyProgram>
<ValueA>A value.</ValueA>
<ValueB>B value.</ValueB>
<ValueC>C value.</ValueC>
</MyProgram>
But I'd like to have this:
<MyProgram Version="1.0">
<ValueA>A value.</ValueA>
<ValueB>B value.</ValueB>
<ValueC>C value.</ValueC>
</MyProgram>
Thanks!
First
Modify your XML to be like this, assuming you have a concrete class and not an anonymous one
[XmlRoot("MyProgram")]//sepcifies the name of the root element
public class MyProgram
{
[XmlAttribute("Version")]//name not required unless you want to change output to something different
public string Version{get;set;}
[XmlElement("ValueA")]//again, name not required if the name is the same
public ValueA ValueA{get;set;}
....
}
Then creating the MyProgram class as you specified will give you the desired output, also note that you might need to add XML tags to the ValueA/B/C classes if the end result is not as desired.
The second
Serialize the XML data to a string, and use simple Regex/String manipulation to insert the desired value and then save the string to the desired location
Third
You can either use XElement to query your XML string /Create it and then set the Version attribute
XElement x = XElement.Load("Your XML location");
var yourRoot = x.Descendants("MyProgram").FirstOrDefaul();
yourRoot.SetAttributeValue("Version","1.0");
yourRoot.Save("Your XML location");
To start, I am constrained to .NET 2.0 so LINQ is not an option for me (though I would be curious to see a LINQ solution as fodder for pushing to move to .NET 3.5 for the project if it is easy).
I have an XSD that is turned into a set of C# classes via xsd.exe at build time. At runtime, an XML file is loaded and deserialized into the C# classes (validation occurs at this time). I need to then turn that in-memory configuration object (including the default values that were populated during import of the XML file) into a dictionary of key value pairs.
I would like the dictionary key to be a dot separated path to the value. Attribute values and element text would be considered values, everything else along the way a key into that.
As an example, imagine the following XML file:
<rootNode>
<foo enabled="true"/>
<bar enabled="false" myAttribute="5.6">
<baz>Some Text</baz>
<baz>Some other text.</baz>
</bar>
</rootNode>
would turn into a dictionary with keys like:
"rootNode.foo.enabled" = (Boolean)true
"rootNode.bar.enabled" = (Boolean)false
"rootNode.bar.myAttribute" = (Float)5.6
"rootNode.bar.baz" = List<String> { "Some Text", "Some other text." }
Things of note are that rootNode is left off not because it is special but because it had no text or attributes. Also, the dictionary is a dictionary of objects which are typed appropriately (this is already done in deserialization, which is one of the reasons I would like to work with the C# object rather than the XML directly).
Interestingly, the objects created by xsd.exe are already really close to the form I want. The class names are things like rootNodeFoo with a float field on it called myAttribute.
One of the things I have considered but am not sure how to go about are using reflection to iterate over the object tree and using the names of the classes of each object to figure out the name of the node (I may have to tweak the casing a bit). The problem with this is that it feels like the wrong solution since I already have access to a deserializer that should be able to do all of that for me and much faster.
Another option would be using XSLT to serialize the data directly to a format that is how I want. The problem here is that my XSLT knowledge is limited and I believe (correct me if I am wrong) I will lose typing on the way (everything will be a string) so I will have to essentially deserialize once again by hand to get the types back out (and this time without XSD validation that I get when I use the .NET deserializer).
In case it matters, the calls I am using to get the configuration object populated from an XML file is something like this:
var rootNode = new XmlRootAttribute();
rootNode.ElementName = "rootNode";
rootNode.Namespace = "urn:myNamespace";
var serializer = new XmlSerializer(typeof(rootNode), rootNode);
using (var reader = new StringReader(xmlString))
{
var deserializedObject = (rootNode)serializer.Deserialize(reader);
}
First observation: using the object graph is not the best place to start to generate a dot representation. You're talking about nodes which have names and are in a well-defined hierarchy and you want to produce some kind of dot notation from it; the xml DOM seems to be the best place to do this.
There are a few problems with the way you describe the problem.
The first is in the strategy when it comes to handling multiple elements of the same name. You've dodged the problem in your example by making that dictionary value actually a list, but suppose your xml looked like this:
<rootNode>
<foo enabled="true">
<bar enabled="false" myAttribute="5.6" />
<bar enabled="true" myAttribute="3.4" />
</foo>
</rootNode>
Besides foo.enabled = (Boolean)true which should be fairly obvious, what dictionary keys do you propose for the two myAttribute leaves? Or would you have a single entry, foo.bar.myAttribute = List<float> {5.6, 3.4}? So, problem #1, there's no unambiguous way to deal with multiple similarly-named non-leaf nodes.
The second problem is in selecting a data type to do the final conversion at leaf nodes (i.e. attribute or element values). If you're writing to a Dictionary<string, object>, you will probably want to select a type based on the Schema simple type of the element/attribute being read. I don't know how to do that, but suggest looking up the various uses of the System.Convert class.
Assuming for the moment that problem #1 won't surface, and that you're ok with a Dictionary<string, string> implementation, here's some code to get you started:
static void Main(string[] args)
{
var xml = #"
<rootNode>
<foo enabled=""true"">
<bar enabled=""false"" myAttribute=""5.6"" />
<baz>Text!</baz>
</foo>
</rootNode>
";
var document = new XmlDocument();
document.LoadXml(xml);
var retVal = new Dictionary<string, string>();
Go(retVal, document.DocumentElement, new List<string>());
}
private static void Go(Dictionary<string, string> theDict, XmlElement start, List<string> keyTokens)
{
// Process simple content
var textNode = start.ChildNodes.OfType<XmlText>().SingleOrDefault();
if (textNode != null)
{
theDict[string.Join(".", keyTokens.ToArray())] = textNode.Value;
}
// Process attributes
foreach (XmlAttribute att in start.Attributes)
{
theDict[string.Join(".", keyTokens.ToArray()) + "." + att.Name] = att.Value;
}
// Process child nodes
foreach (var childNode in start.ChildNodes.OfType<XmlElement>())
{
Go(theDict, childNode, new List<string>(keyTokens) { childNode.Name }); // shorthand for .Add
}
}
And here's the result:
One approach would be to implement a customer formatter and slot it into the standard serialization pattern, create a class that implements IFormatter i.e. MyDotFormatter
http://msdn.microsoft.com/en-us/library/system.runtime.serialization.iformatter.aspx
then implement as below
Stream stream = File.Open(filename, FileMode.Create);
MyDotFormatter dotFormatter = new MyDotFormatter();
Console.WriteLine("Writing Object Information");
try
{
dotFormatter.Serialize(stream, objectToSerialize);
}
catch (SerializationException ex)
{
Console.WriteLine("Exception for Serialization data : " + ex.Message);
throw;
}
finally
{
stream.Close();
Console.WriteLine("successfully wrote object information");
}
When i am doing Deserialize of xml i am getting "There is an error in XML document (1, 41)." . Can anyone tell me about what is the issue is all about.
public static T DeserializeFromXml<T>(string xml)
{
T result;
XmlSerializer ser = new XmlSerializer(typeof(T));
using (TextReader tr = new StringReader(xml))
{
result = (T)ser.Deserialize(tr);
}
return result;
}
I use this function to do it.
<?xml version='1.0' encoding='utf-16'?>
<Message>
<FirstName>Hunt</FirstName>
<LastName>DAvid</LastName>
</Message>
Ensure your Message class looks like below:
[Serializable, XmlRoot("Message")]
public class Message
{
public string FirstName { get; set; }
public string LastName { get; set; }
}
This works for me fine:
string xml = File.ReadAllText("c:\\Message.xml");
var result = DeserializeFromXml<Message>(xml);
MSDN, XmlRoot.ElementName:
The name of the XML root element that is generated and recognized in
an XML-document instance. The default is the name of the serialized
class.
So it might be your class name is not Message and this is why deserializer was not able find it using default behaviour.
Agreed with the answer from sll, but experienced another hurdle which was having specified a namespace in the attributes, when receiving the return xml that namespace wasn't included and thus failed finding the class.
i had to find a workaround to specifying the namespace in the attribute and it worked.
ie.
[Serializable()]
[XmlRoot("Patient", Namespace = "http://www.xxxx.org/TargetNamespace")]
public class Patient
generated
<Patient xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns="http://www.xxxx.org/TargetNamespace">
but I had to change it to
[Serializable()]
[XmlRoot("Patient")]
public class Patient
which generated to
<Patient xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
This solved my problem, hope it helps someone else.
First check the variables declared using proper Datatypes.
I had a same problem then I have checked, by mistake I declared SAPUser as int datatype so that the error occurred.
One more thing XML file stores its data using concept like array but its first index starts having +1.
e.g. if error is in(7,2) then
check for 6th line always.....
I had the same thing. All came down to a "d" instead of a "D" in a tag name in the schema.
In my case I had a float value expected where xml had a null value so be sure to search for float and int data type in your xsd map
On a WEC7 project I'm working on, I got a similar error. The file I was serializing in was serialized out from an array of objects, so I figured the XML was fine. Also, I have had this working for a few previous classes, so it was quite a puzzle.
Then I noticed in my earlier work that every class that I was serializing/deserializing had a default constructor. That was missing in my failed case so I added it and and voila... it worked fine.
I seem to remember reading somewhere that this was required. I guess it is.
I'm using
List<EFacebook> facebooks = BFacebook.ReadFacebookFriends(user.EProviders
.Where(i => i.ProviderType == EProvider.EnumProviderType.Facebook)
.First().Token);
StringBuilder xml = new StringBuilder();
foreach (EFacebook i in facebooks)
{
xml.AppendFormat("<id>{0}</id>", i.id);
}
Can anybody suggest a better code to serialize each i.id into an XML string?
Edit:
The facebook object has close to 20 properties. If I XmlSerializer all 20 properties are serialized into the XML. I just need the id column.
You might want to take a look at XML Serialization already built into the .NET Framework.
You can use code similar to the following to serialize the object:
MySerializableClass myObject = new MySerializableClass();
// Insert code to set properties and fields of the object.
XmlSerializer mySerializer = new
XmlSerializer(typeof(MySerializableClass));
// To write to a file, create a StreamWriter object.
StreamWriter myWriter = new StreamWriter("myFileName.xml");
mySerializer.Serialize(myWriter, myObject);
myWriter.Close();
See: How to serialize
You can flag properties to be ignored using the XmlIgnore attribute as shown below:
public class Group
{
// The XmlSerializer ignores this field.
[XmlIgnore]
public string Comment;
// The XmlSerializer serializes this field.
public string GroupName;
}
See XMLIgnore
XmlSerializer xs = new XmlSerializer(typeof(int[]));
xs.Serialize(stream,facebooks.Select(x=>x.id).ToArray())
I would use Linq To Xml to create the Xml Tree structure
See for an example: Linq To Xml
If you're just saving the id values into an xml string then I wouldn't say that you're 'serializing' the objects.
Whenever working with XML it's much nicer to use an XML library than to wrangle with text. One obvious reason is that it guarantees your XML will be well-formed.
I'd do something like this:
List<EFacebook> facebooks = GetFriends();
var facebookIds = facebooks.Select(f => new XElement("id", f.id));
var facebookXml = new XElement("facebookIds", facebookIds);
Which will give you XML like
<facebookIds>
<id>1</id>
<id>2</id>
</facebookIds>
I'm calling a "web service" that gives me as an xml response an invalid node, so when I try to deserialize it, it throws an exception.
I'm using the XmlSerializer class, like so:
internal class Response<T>
{
public Response(byte[] xml)
{
XmlSerializer s = new XmlSerializer(typeof(T));
XmlReader reader = XmlReader.Create(new MemoryStream(xml));
if (s.CanDeserialize(reader))
this.ActualResponse = (T)s.Deserialize(reader);
}
public T ActualResponse { get; private set; }
}
and the node I'm having trouble with looks something like this:
<autorizacion>FALSE</autorizacion>
The exception I get is
System.InvalidOperationException:
There is an error in XML document (7,
35). ---> System.FormatException: The
string 'FALSE' is not a valid Boolean
value..
Which is obvious.
The question is, how can I deserialize it without having to iterate through all nodes, building my response entity by hand? Is there a way?
I don't have control over the server
Quickest way seems to be to change the parameter of setAutorizacion(boolean) to setAutorizacion(String), then convert to a boolean in the setter. Also, document what you did and why you did it both in that setter and in more high-level documentation.
you can use an Xsl to reformat your xml before the deserialization
EDIT
for xsl transform with c#: http://www.csharpfriends.com/Articles/getArticle.aspx?articleID=63
your xsl should contains something like that
<xsl:template match="autorizacion">
<autorizacion><xsl:value-of select="concat(upper-case(substring(current(),1,1)), substring(current(),2))" />
</autorizacion>
</xsl:template>
you'll have to format it without validating it to your schema. format the values that are wrong and revalidate it with schema.