Develop Reference

C# XML Serialization - Custom ordering based on data [duplicate]

I have the following test XML string:
<?xml version="1.0" encoding="UTF-8"?>
<test id="myid">
<b>b1</b>
<a>a2</a>
<a>a1</a>
<b>b2</b>
</test>
which I deserialize using this class:
[XmlRoot(ElementName = "test")]
public class Test
{
[XmlElement(ElementName = "a")]
public List<string> A { get; set; }
[XmlElement(ElementName = "b")]
public List<string> B { get; set; }
[XmlAttribute(AttributeName = "id")]
public string Id { get; set; }
}
If I'm now going to serialize the object the result will be:
<?xml version="1.0" encoding="utf-16"?>
<test xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" id="myid">
<a>a2</a>
<a>a1</a>
<b>b1</b>
<b>b2</b>
</test>
Is there a way to keep the initial sort order?
I guess I can't use [XmlElementAttribute(Order = x)] cause the order shouldn't be hardcoded but identically with the initial xml.
I thought about adding an order property to my lists like that
[XmlRoot(ElementName="a")]
public class A
{
[XmlAttribute(AttributeName="order")]
public string Order { get; set; }
[XmlText]
public string Text { get; set; }
}
[XmlRoot(ElementName="b")]
public class B
{
[XmlAttribute(AttributeName="order")]
public string Order { get; set; }
[XmlText]
public string Text { get; set; }
}
[XmlRoot(ElementName="test")]
public class Test
{
[XmlElement(ElementName="a")]
public List<A> A { get; set; }
[XmlElement(ElementName="b")]
public List<B> B { get; set; }
[XmlAttribute(AttributeName="id")]
public string Id { get; set; }
}
but I don't know how to order by them when serializing.

You can do this with XmlSerializer by using a single collection to capture both the <a> and <b> elements and applying the [XmlElement(Name, Type = typeof(...))] attribute to it multiple times, once for each desired element name. Because you are using a single collection to deserialize both elements, the order is automatically preserved. However, to make this work, XmlSerializer must be able to determine the correct element name when re-serializing. There are two approaches to accomplish this, as documented in Choice Element Binding Support:
If the collection contains polymorphic items, the element name can be mapped to the concrete item type by using the [XmlElementAttribute(String, Type)] constructor. For instance, if you have a sequence of elements that might be strings or integers like so:
<Things>
<string>Hello</string>
<int>999</int>
</Things>
This can be bound to a collection as follows:
public class Things
{
[XmlElement(Type = typeof(string)),
XmlElement(Type = typeof(int))]
public List<object> StringsAndInts { get; set; }
}
If the collection contains only a single type of item, the element name can be encoded in an associated array of enum values, where the enum names correspond to the element names and the array itself is identified via the [XmlChoiceIdentifierAttribute] attribute.
For details, see the documentation examples.
I find option #1 easier to work with than option #2. Using this approach, the following model will deserialize and re-serialize your XML while successfully preserving the order of the <a> and <b> elements:
public abstract class StringElementBase
{
[XmlText]
public string Text { get; set; }
public static implicit operator string(StringElementBase element)
{
return element == null ? null : element.Text;
}
}
public sealed class A : StringElementBase
{
}
public sealed class B : StringElementBase
{
}
[XmlRoot(ElementName = "test")]
public class Test
{
[XmlElement("a", Type = typeof(A))]
[XmlElement("b", Type = typeof(B))]
public List<StringElementBase> Items { get; } = new List<StringElementBase>();
[XmlIgnore]
// For convenience, enumerate through the string values of the items.
public IEnumerable<string> ItemValues { get { return Items.Select(i => (string)i); } }
[XmlAttribute(AttributeName = "id")]
public string Id { get; set; }
}
Working .Net fiddle.
For more examples of using [XmlChoiceIdentifier] to deserialize a sequence of elements with different names to c# objects of the same type, see for instance here or here.

No, basically; XmlSerializer doesn't support that. If you want to use that option you'd need to write it manually with XDocument or XmlDocument or XmlReader.

XML Deserialization is returning null for an xml attribute but is returning the correct Xml text

Below I have the following objects:
[XmlRoot("ExampleObject")]
public class ExampleObject
{
[XmlElement("element1")]
public long element1{ get; set; }
[XmlElement("element2")]
public string element2{ get; set; }
[XmlElement("element3")]
public string element3{ get; set; }
[XmlElement("element4")]
public ExampleObject2 element4{ get; set; }
}
[Serializable()]
public class ExampleObject2
{
[XmlText]
public string Value { get; set; }
[XmlAttribute]
public string Id { get; set; }
[XmlAttribute]
public string Ref { get; set; }
}
and the following XML:
<c:ExampleObject>
<c:element1>
...
</c:element1>
<c:element2>
...
</c:element2>
<c:element3>
...
</c:element3>
<c:element4 z:Ref="953" i:nil="true"/> <!--- or this <c:Status z:Id="953">...</c:element4> -->
</ExampleObject>
I am able to get my object perfectly fine, except for element4 where inside ExampleObject2 my Id and Ref are null while Value has the correct value inside.
I want it so that ExampleObject2 always has a Value, it is either null or the value it is supposed to has. I want Id and Ref to be the value of the attribute inside element4 depending on if element4 has that attribute or not.
I was wondering if there's something I am missing?
I have also tried to add specific names for the XMLElement tags like this [XmlAttribute("Id")] but that doesn't seem to do anything. I have also tried including the namespace like this [XmlAttribute("z:Id")] but that causes an error.
I have also tried putting IsNullable within the element4 XMLElement tag but for the XML elements that have nil="true" but it makes the entire ExampleObject2 null which isn't exactly what I am looking for.
I've already looked at the following question and have come to the same issue.
how to deserialize an xml node with a value and an attribute using asp.net serialization
Thank you

I ended up fixing it with the Nil property like #GeorgeBirbilis said and used it with #dbc's solution to manually naming the namespaces the z: and c: are from. I was only doing one or the other, not both combined. Thank you guys for the help.
[XmlAttribute(Namespace = "http://schemas.microsoft.com/2003/10/Serialization/")]
public int Id { get; set; }
[XmlAttribute(Namespace = "http://schemas.microsoft.com/2003/10/Serialization/")]
public int Ref { get; set; }
private bool _nil;
[XmlAttribute("nil", Namespace = "http://www.w3.org/2001/XMLSchema-instance")]
public bool Nil
{
get
{
return _nil;
}
set
{
_nil = value;
}
}

Deserialize xml Elements with same Attributes

I am deserializing a xml file with an attribute and a xml text. The problem is that those elements have the same attributes. So I am always getting the error that I must not have two identical TypeNames in XmlType.
My xml:
<group_id xsi:type="xsd:int">1</group_id>
<name xsi:type="xsd:int">myNameView</name>
And my C#:
[XmlType(AnonymousType = true, Namespace = "http://www.w3.org/2001/XMLSchema", TypeName = "int")]
[XmlRoot(ElementName = "group_id")]
public class Group_id
{
[XmlAttribute(AttributeName = "type", Namespace = "http://www.w3.org/2001/XMLSchema-instance")]
public string Type { get; set; }
[XmlText]
public string Text { get; set; }
}
[XmlType(AnonymousType = true, Namespace = "http://www.w3.org/2001/XMLSchema", TypeName = "int")]
[XmlRoot(ElementName = "name")]
public class Name
{
[XmlAttribute(AttributeName = "type", Namespace = "http://www.w3.org/2001/XMLSchema-instance")]
public string Type { get; set; }
[XmlText]
public string Text { get; set; }
}
The problem is the TypeName in XmlType Attribute. If I name only one element with a TypeName it is correctly deserialized.

XmlSerializer handles types for you, based on the xsi:type attribute. By trying to handle these yourself, you're causing it some grief.
If you declare your elements as object, then the serializer will use the type attributes to determine how to deserialize the values. Note as your example is just a fragment, I've assumed root element called root:
[XmlRoot("root")]
public class Root
{
[XmlElement("group_id")]
public object GroupId { get; set; }
[XmlElement("name")]
public object Name { get; set; }
}
Now when you deserialize your example, you'd actually get an exception (as myNameView isn't an integer). You can fix by either changing the type to xsd:string or changing the value to a valid integer.
In the event the XML was valid, you'd see that the types deserialized map directly to the type attributes. See this fiddle for a working demo.

XML - Deserialize property as Xml sub-tree

When I'm deserializing a xml string, I need to save a XElement outerXml on a string property called prop2.
My XML:
<MyObj>
<prop1>something</prop1>
<prop2>
<RSAKeyValue>
<Modulus>...</Modulus>
<Exponent>...</Exponent>
</RSAKeyValue>
</prop2>
<prop3></prop3>
</MyObj>
My object:
public class MyObj
{
[XmlElement("prop1")]
public string prop1 { get; set; }
[XmlText]
public string prop2 { get; set; }
[XmlElement(ElementName = "prop3", IsNullable = true)]
public string prop3 { get; set; }
}
I'm deserializing using XmlSerializer, like this:
var serializer = new XmlSerializer(typeof(T));
return (T)serializer.Deserialize(new StringReader(myXmlString));
I tried to use [XmlText] to save XML text in prop2 But I'm only getting null.
What I need to do to save <RSAKeyValue><Modulus>...</Modulus><Exponent>...</Exponent></RSAKeyValue> like text in prop2?

XmlText will give value as XML encoded as text (">prop2<...") see XmlTextAttribute
By default, the XmlSerializer serializes a class member as an XML element. However, if you apply the XmlTextAttribute to a member, the XmlSerializer translates its value into XML text. This means that the value is encoded into the content of an XML element.
One possible solution - use XmlNode as type of the property:
public class MyObj
{
[XmlElement("prop1")]
public string prop1 { get; set; }
public XmlNode prop2 { get; set; }
[XmlElement(ElementName = "prop3", IsNullable = true)]
public string prop3 { get; set; }
}
var r = (MyObj)serializer.Deserialize(new StringReader(myXmlString));
Console.WriteLine(r.prop2.OuterXml);
Alternatively you may make whole object implement custom Xml serialization or have custom type that matches XML (to read normally) and have additional property to represent that object as XML string.

C#, XML to Class DeSerialization Question

I have the following XML snippet:
<dmFiles>
−
<dmFile dmUpFileGuid="" dmFileDescr="testcase01.pdf" dmFileMetaType="main" dmFileGuid="" dmMimeType="pdf" dmFormat="">
If I create a strongly typed C# class with string properties for the dmFile attributes (eg: dmFileDescr), how can I ensure these attributes will serialize to properties in my c# class?

By using Xml attributes on your class member. Use [XmlAttribute("name")].
Your implementation would look like this:
[XmlRoot("dmFile")]
public class DmFile
{
[XmlAttribute("dmUpFileGuid")]
public String UpFileGuid { get;set; }
...
}

Try this:
[Serializable]
[XmlRoot(ElementName="dmFile")]
public class File
{
[XmlAttribute(AttributeName="dmUpFileGuid")]
public string UploadGuid { get; set; }
[XmlAttribute(AttributeName = "dmFileDescr")]
public string Description { get; set; }
[XmlAttribute(AttributeName = "dmFileMetaType")]
public string MetaType { get; set; }
[XmlAttribute(AttributeName = "dmFileGuid")]
public string FileGuid { get; set; }
[XmlAttribute(AttributeName = "dmMimeType")]
public string MimeType { get; set; }
[XmlAttribute(AttributeName = "dmFormat")]
public string Format { get; set; }
}
And deserialize your XML as follow:
XmlSerializer s = new XmlSerializer(typeof(File));
File file = s.Deserialize(new StringReader(#"<dmFile ... />")) as File;

You can (de)serialize from/to XML with the XmlSerializer and marking up the target class with Attributes provided for Xml-Serialization.
Mark your public properties with the correct attribute. It should be XmlAttributeAttribute.
The enclosing class must map on the dmFile-Element (XmlRootAttribute)
If the property is called differently, or the class is called differently than the XML element, you need to specify the XML-Element/Attribute name.