I am trying to Deserialize the following XML:
<Test><string name="Name">Test name</string><string name="Description">Some fake description.</string></Test>
Into the following class.
[XmlRoot("Test")]
public class Test
{
[XmlElement("string")]
public string Name;
[XmlElement("string")]
public string Description;
}
Using the code I am doing it with.
var xml = #"<Test><string name=""Name"">Test name</string><string name=""Description"">Some fake description.</string></Test>";
XmlReader reader = new XmlTextReader(new StringReader(xml));
XmlSerializer serializer = new XmlSerializer(typeof(Test));
serializer.Deserialize(reader);
When I run this I get an InvalidOperationException with the message
There was an error reflecting type 'Test'.
If I comment out the Description property, it works. I can get the attribute value or the text, but I can't get just the XmlText with the element is string and the "name" Attribute has a specific value.
Is this even possible without using LINQ?
As per my comment:
Certainly won't be able to do it without changing something. You're
telling .NET that Description is an element when it's an attribute of
the 'string' element. Use LINQ
Here is an example of the LINQ, it's quite straightforward to extend and decouples your XML from your class (which is often a good thing!).
var xml = #"<Test><string name=""Name"">Test name</string><string name=""Description"">Some fake description.</string></Test>";
var xdoc = XDocument.Parse(xml);
var output = from test in xdoc.Elements("Test")
let strings = test.Elements("string").ToDictionary(e => e.Attribute("name").Value, e => e.Value)
select new Test () { Name = strings["Name"],
Description = strings["Description"] };
The reason is that you are not using XmlElement as intended, the element name "string" must be unique the class. The "name" attribute is not taken into account.
So, in summary, it is not possible to deserialize that xml document automatically, you would need to implement the deserialization method by yourself.
For that you would need:
public class Foo {
[XmlAttribute("name")]
public string Name {get;set;}
[XmlText]
public string Value {get;set;}
}
Then, in the parent type:
[XmlRoot("Test")]
public class Test
{
[XmlElement("string")]
public List<Foo> Items {get;set;}
}
This it the only way you can process that shape XML unless you use IXmlSerializable (very hard).
Related
I have an entity as below
public class Vehicle{
public int VehicleId {get;set;};
public string Make {get;set;};
public string Model{get;set;}
}
I wanted to serialize as below
<Vehicle>
<VehicleId AppliesTo="C1">1244</VehicleId>
<Make AppliesTo="Common" >HXV</Make>
<Model AppliesTo="C2">34-34</Model>
</Vehicle>
I have around 100 properties like this in Vehicle class, for each vehicle property I wanted to attach a metadata ApplieTo which will be helpful to downstream systems. AppliesTo attribute is static and its value is defined at the design time. Now How can I attach AppliesTo metadata to each property and inturn get serialized to XML?
You can use XElement from System.Xml.Linq to achieve this. As your data is static you can assign them easily. Sample code below -
XElement data= new XElement("Vehicle",
new XElement("VehicleId", new XAttribute("AppliesTo", "C1"),"1244"),
new XElement("Make", new XAttribute("AppliesTo", "Common"), "HXV"),
new XElement("Model", new XAttribute("AppliesTo", "C2"), "34 - 34")
);
//OUTPUT
<Vehicle>
<VehicleId AppliesTo="C1">1244</VehicleId>
<Make AppliesTo="Common">HXV</Make>
<Model AppliesTo="C2">34 - 34</Model>
</Vehicle>
If you are not interested in System.Xml.Linq then you have another option of XmlSerializer class. For that you need yo define separate classes for each property of vehicle. Below is the sample code and you can extend the same for Make and Model -
[XmlRoot(ElementName = "VehicleId")]
public class VehicleId
{
[XmlAttribute(AttributeName = "AppliesTo")]
public string AppliesTo { get; set; }
[XmlText]
public string Text { get; set; }
}
[XmlRoot(ElementName = "Vehicle")]
public class Vehicle
{
[XmlElement(ElementName = "VehicleId")]
public VehicleId VehicleId { get; set; }
//Add other properties here
}
Then create test data and use XmlSerializer class to construct XML -
Vehicle vehicle = new Vehicle
{
VehicleId = new VehicleId
{
Text = "1244",
AppliesTo = "C1",
}
};
XmlSerializer testData = new XmlSerializer(typeof(Vehicle));
var xml = "";
using (var sww = new StringWriter())
{
using (XmlWriter writer = XmlWriter.Create(sww))
{
testData.Serialize(writer, vehicle);
xml = sww.ToString(); // XML
}
}
It is not easy or ideal to use the default .NET XML serializer (System.Xml.Serialization.XmlSerializer) in the way you want, but it's possible. This answer shows how to create a class structure to hold both your main data and the metadata, then use XmlAttributeAttribute to mark a property so it gets serialized as an XML attribute.
Assumptions:
There are a number of unknowns about your intended implementation, such as:
The XML serializer you want to use (default for .NET?)
The mechanism to inject 'AppliesTo' (attribute?)
Do you care about deserialization?
This answer assumes the default .NET serializer, that deserialization matters, and that you don't care about the exact method of injecting your metadata.
Key concepts:
A generic class to hold both our main property value and the metadata (see PropertyWithAppliesTo<T>)
Using XmlAttributeAttribute on the generic class' metadata, so it is written as an XML attribute on the parent property
Using XmlTextAttribute on the generic class' main data, so it is written as the Xml text of the parent property (and not as a sub-property)
Including two properties on the main type being serialized (in this case Vehicle) for every value you want serialized: one of the new generic type that gets serialized with metadata, and one of the original type marked with XmlIgnoreAttribute that provides 'expected' access to the property's value
Using the XmlElementAttribute to change the name of the serialized property (so it matches the expected name)
Code:
using System;
using System.IO;
using System.Xml.Serialization;
namespace SomeNamespace
{
public class Program
{
static void Main()
{
var serializer = new XmlSerializer(typeof(Vehicle));
string s;
var vehicle = new Vehicle { VehicleId = 1244 };
//serialize
using (var writer = new StringWriter())
{
serializer.Serialize(writer, vehicle);
s = writer.ToString();
Console.WriteLine(s);
}
// edit the serialized string to test deserialization
s = s.Replace("Common", "C1");
//deserialize
using (var reader = new StringReader(s))
{
vehicle = (Vehicle)serializer.Deserialize(reader);
Console.WriteLine($"AppliesTo attribute for VehicleId: {vehicle.VehicleIdMeta.AppliesTo}");
}
}
}
public class Vehicle
{
[XmlElement(ElementName = "VehicleId")] // renames to remove the 'Meta' string
public PropertyWithAppliesTo<int> VehicleIdMeta { get; set; } = new PropertyWithAppliesTo<int>("Common");
[XmlIgnore] // this value isn't serialized, but the property here for easy syntax
public int VehicleId
{
get { return VehicleIdMeta.Value; }
set { VehicleIdMeta.Value = value; }
}
}
public class PropertyWithAppliesTo<T>
{
[XmlAttribute] // tells serializer this should be an attribute on this element, not a property
public string AppliesTo { get; set; } = string.Empty;
[XmlText] // tells serializer to not write this as a property, but as the main XML text
public T Value { get; set; } = default;
public PropertyWithAppliesTo() : this(string.Empty) { }
public PropertyWithAppliesTo(string appliesTo) : this(appliesTo, default) { }
public PropertyWithAppliesTo(string appliesTo, T initialValue)
{
AppliesTo = appliesTo;
Value = initialValue;
}
}
}
When run, the string s will look like:
<?xml version=\"1.0\" encoding=\"utf-16\"?>
<Vehicle xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\">
<VehicleId AppliesTo="Common">1244</VehicleId>
</Vehicle>
Other Notes:
You can see how to add more properties to Vehicle: add a property of type PropertyWithAppliesTo<T> marked with XmlElement to give it the name you want, and then a property of type T marked with XmlIgnore that wraps around the Value you want.
You can control the value of AppliesTo by changing the input to the constructor of PropertyWithAppliesTo<T> and giving it a different metadata string.
If you don't want consumers of your library to see the 'meta' properties in IntelliSense, you can use the EditorBrowsableAttribute. It won't hide things from you when using the source and a project reference; it's only hidden when referencing the compiled dll.
This is admittedly an annoying way to add properties to a class. But if you want to use the default .NET XML serializer, this is a way to achieve the XML you want.
I have two different XML documents. Their structure is almost identical, but they have some different elements.
I would like to deserialize the incoming documents in to one class which is a superset of both classes. There is no need to ever serialize the class, I only need to deserialize the documents.
The XML document types have a different root element, lets say the root of the first is <CLASSA>and the other is <CLASSB>.
I am looking for something like this, where both <CLASSA> and <CLASSB> xml documents are mapped to ClassAandB:
[XmlRoot(ElementName="CLASSA,CLASSB")]
public class ClassAandB {
[XmlElement(ElementName="syntaxid")]
public Syntaxid Syntaxid{ get; set; }
[XmlElement(ElementName="email")]
public Email Email { get; set; }
[XmlElement(ElementName="envelope")]
public Envelope Envelope { get; set; }
[XmlElement(ElementName="header")]
public Header Header { get; set; }
}
I can then find out which of the two types it is by reading the Syntaxid property. This helps me because a lot of the processing is the same for both types.
Any suggestions how to do this?
Since the xml root element name might depend on the content of the xml document, you'll have to configure the XmlSerializer at runtime with this xml root element name to be used.
In this case, there's no need anymore to apply an XmlRootAttribute.
This can be done via the constructor overload accepting an XmlRootAttribute argument, via which you pass the root element name.
public XmlSerializer (Type type, System.Xml.Serialization.XmlRootAttribute root);
You might know the root element name in front eg. depending on the source of the xml document, or you might discover it at runtime from the xml document itself.
The following from the example below shows how the xml root element name gets set.
String rootName = "CLASSA"; // "CLASSB"
var serializer = new XmlSerializer(typeof(ClassAandB), new XmlRootAttribute(rootName));
An simplified example using an XmlReader as source and retrieving the root xml element name from the content.
public class ClassAandB
{
[XmlElement(ElementName="syntaxid")]
public String Syntaxid{ get; set; }
[XmlElement(ElementName="email")]
public String Email { get; set; }
[XmlElement(ElementName="header")]
public String Header { get; set; }
}
var classA = Deserialize(XmlReader.Create(
new StringReader("<CLASSA><syntaxid>A</syntaxid></CLASSA>"))
);
Console.WriteLine(classA.Syntaxid); // A
var classB = Deserialize(
XmlReader.Create(new StringReader("<CLASSB><syntaxid>B</syntaxid></CLASSB>"))
);
Console.WriteLine(classB.Syntaxid); // B
public static ClassAandB Deserialize(XmlReader reader)
{
reader.MoveToContent();
string rootName = reader.Name;
var serializer = new XmlSerializer(typeof(ClassAandB),
new XmlRootAttribute(rootName)
);
var deserialized = serializer.Deserialize(reader) as ClassAandB;
return deserialized;
}
I suggest you to remove XmlRoot attribute and use:
var doc = new XmlDocument();
doc.Load("file.xml");
XmlElement root = xmlDoc.DocumentElement;
var serializer = new XmlSerializer(typeof(ClassAandB), new XmlRootAttribute(root.ToString()));
I have an List<AnimalsEnum> Foo property in a class that I'm serializing to XML with RestSharp for the body of a request. I'd like the output to be:
<rootNode>
... existing content...
<Foo>Elephant</Foo>
<Foo>Tiger</Foo>
.... more content
Instead, for the relevant serialisation part, I have
<Foo>
<AnimalsEnum />
<AnimalsEnum />
</Foo>
I'd like to convert the enum values to strings and remove the container element that is automatically added. Is this possible with RestSharp? I thought it may be possible with attributes, but apparently not. Am I going to have to wrangle this output myself with a custom serialiser?
Code is difficult to post, but keeping with the example:
class Bar
{
public string Name{get;set;}
public List<AnimalsEnum> Foo{get;set;}
public enum AnimalsEnum {Tiger,Elephant,Monkey}
}
and to serialize into a request
var req = new RestSharp.RestRequest(RestSharp.Method.POST);
req.RequestFormat = RestSharp.DataFormat.Xml;
req.AddQueryParameter("REST-PAYLOAD", "");
req.AddXmlBody(myBar);
You can use the built-in DotNetXmlSerializer of RestSharp to make Microsoft's XmlSerializer do the actual serialization. Then you can use XML serialization attributes to specify that the List<AnimalsEnum> of Bar should be serialized without an outer container element by applying [XmlElement]:
public class Bar
{
public string Name { get; set; }
[System.Xml.Serialization.XmlElement]
public List<AnimalsEnum> Foo { get; set; }
public enum AnimalsEnum { Tiger, Elephant, Monkey }
}
Then, when making the request, do:
var req = new RestSharp.RestRequest(RestSharp.Method.POST);
// Use XmlSerializer to serialize Bar
req.XmlSerializer = new RestSharp.Serializers.DotNetXmlSerializer();
req.RequestFormat = RestSharp.DataFormat.Xml;
req.AddQueryParameter("REST-PAYLOAD", "");
req.AddXmlBody(myBar);
Note that Bar must be public because XmlSerializer can only serialize public types.
The XML I am getting is provided by an outside source so I don't have the ability to easily reformat it. I would like to use xml attributes on my entities instead of having to write a linq query that knows how the XML and entity is formatted. Here is an example:
<?xml version="1.0"?>
<TERMS>
<TERM>
<ID>2013-2</ID>
<DESC>Spring 2013</DESC>
</TERM>
<TERM>
<ID>2013-3</ID>
<DESC>Summer 2013 Jun&Jul</DESC>
</TERM>
</TERMS>
I know the the XMLSerializer expects ArrayOfTerm instead of TERMS for example, but that I can tweak my entity to use a different element name with the xml attributes such as this:
public class TermData
{
[XmlArray("TERMS")]
[XmlArrayItem("TERM")]
public List<Term> terms;
}
public class Term
{
[XmlElement("ID")]
public string id;
[XmlElement("DESC")]
public string desc;
}
and I am deserializing the data like so:
TermData data;
XmlSerializer serializer = new XmlSerializer(typeof(TermData));
using (StringReader reader = new StringReader(xml))
{
data = (TermData)serializer.Deserialize(reader);
}
return View(data.terms);
The problem I am facing is that TERMS is the root and the array itself. If the XML were to have a root element that was not the array, I could edit my TermData class like so and it would deserialize correctly (already tested).
[XmlRoot("ROOT")]
public class TermData
{
[XmlArray("TERMS")]
[XmlArrayItem("TERM")]
public List<Term> terms;
}
Note that using TERMS as the XMLRoot does not work. Right now, my code is throwing
InvalidOperationException: There is an error in XML document (2,2).
InnerException: "<TERMS xmlns=" was not expected.
This would lead me to believe that the XML is not formatted correctly, but from my understanding the example I gave is perfectly valid XML.
This would all be trivial if I could edit the source xml, but there could be tons of other responses like this and I need to be able to flex for whatever I might get. What I'm trying to confirm is whether or not the XMLSerializer can support this type of XML structure. I've tested just about everything and can't get it deserialize without editing the XML. It would also be convenient if I didn't have to define a wrapper class (TermData) to hold the list, but this seems to only work if the xml follows the naming conventions for the serializer (ArrayOfTerm, etc).
Maybe you can try :
[XmlRoot("TERMS")]
public class TermData
{
public TermData()
{
terms = new List<Term>();
}
[XmlElement("TERM")]
public List<Term> terms{get;set;}
}
public class Term
{
[XmlElement("ID")]
public string id{get;set;}
[XmlElement("DESC")]
public string desc{get;set;}
}
Hope this will help,
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
C#: Read XML Attribute using XmlDocument
In C# if I were to have the XMLDocument containing:
<Hello>
<Person:"Alan" Saying:"My name is Alan">
</Hello>
Then how might I obtain the single attribute "Saying"? I've found code which works when a single attribute is contained within the "<>" however it does not appear to work where there are multiple attributes inside as is above.
Any help would be much appreciated, I'm rather a C# novice :)
First off, that is not valid xml. I think you want something like
<Hello>
<Person name="Alan" Saying="My name is Alan" />
</Hello>
and the simplest way to get Alan's Saying is
XmlDocument doc = new XmlDocument();
doc.Load("filename.xml");
string saying = doc.SelectSingleNode("//Person[#name=Alan]").Attributes["saying"].Value;
for a more detailed explanation of why that works, see XPath Examples
Your XML should be:
<Hello>
<Person name="Alan" saying="My name is Alan"/>
</Hello>
Your class to deserialize this would be:
public class Hello
{
public Person Person { get; set; }
}
public class Person
{
[XmlAttribute]
public string Name { get; set; }
[XmlAttribute]
public string Saying { get; set; }
}
How to use it:
// Create a new XmlSerializer instance with the type of the test class
XmlSerializer SerializerObj = new XmlSerializer(typeof(Hello));
// load xml into string reader
StringReader reader = new StringReader(yourXmlString);
// Load the object saved above by using the Deserialize function
Hello LoadedObj = (Hello)SerializerObj.Deserialize(reader);
Check out the MSDN article for more info on how to use the XmlSerializer