How do I Deserialize this XML document:
<?xml version="1.0" encoding="utf-8"?>
<Cars>
<Car>
<StockNumber>1020</StockNumber>
<Make>Nissan</Make>
<Model>Sentra</Model>
</Car>
<Car>
<StockNumber>1010</StockNumber>
<Make>Toyota</Make>
<Model>Corolla</Model>
</Car>
<Car>
<StockNumber>1111</StockNumber>
<Make>Honda</Make>
<Model>Accord</Model>
</Car>
</Cars>
I have this:
[Serializable()]
public class Car
{
[System.Xml.Serialization.XmlElementAttribute("StockNumber")]
public string StockNumber{ get; set; }
[System.Xml.Serialization.XmlElementAttribute("Make")]
public string Make{ get; set; }
[System.Xml.Serialization.XmlElementAttribute("Model")]
public string Model{ get; set; }
}
.
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlArrayItem(typeof(Car))]
public Car[] Car { get; set; }
}
.
public class CarSerializer
{
public Cars Deserialize()
{
Cars[] cars = null;
string path = HttpContext.Current.ApplicationInstance.Server.MapPath("~/App_Data/") + "cars.xml";
XmlSerializer serializer = new XmlSerializer(typeof(Cars[]));
StreamReader reader = new StreamReader(path);
reader.ReadToEnd();
cars = (Cars[])serializer.Deserialize(reader);
reader.Close();
return cars;
}
}
that don't seem to work :-(
How about you just save the xml to a file, and use xsd to generate C# classes?
Write the file to disk (I named it foo.xml)
Generate the xsd: xsd foo.xml
Generate the C#: xsd foo.xsd /classes
Et voila - and C# code file that should be able to read the data via XmlSerializer:
XmlSerializer ser = new XmlSerializer(typeof(Cars));
Cars cars;
using (XmlReader reader = XmlReader.Create(path))
{
cars = (Cars) ser.Deserialize(reader);
}
(include the generated foo.cs in the project)
Here's a working version. I changed the XmlElementAttribute labels to XmlElement because in the xml the StockNumber, Make and Model values are elements, not attributes. Also I removed the reader.ReadToEnd(); (that function reads the whole stream and returns a string, so the Deserialize() function couldn't use the reader anymore...the position was at the end of the stream). I also took a few liberties with the naming :).
Here are the classes:
[Serializable()]
public class Car
{
[System.Xml.Serialization.XmlElement("StockNumber")]
public string StockNumber { get; set; }
[System.Xml.Serialization.XmlElement("Make")]
public string Make { get; set; }
[System.Xml.Serialization.XmlElement("Model")]
public string Model { get; set; }
}
[Serializable()]
[System.Xml.Serialization.XmlRoot("CarCollection")]
public class CarCollection
{
[XmlArray("Cars")]
[XmlArrayItem("Car", typeof(Car))]
public Car[] Car { get; set; }
}
The Deserialize function:
CarCollection cars = null;
string path = "cars.xml";
XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
StreamReader reader = new StreamReader(path);
cars = (CarCollection)serializer.Deserialize(reader);
reader.Close();
And the slightly tweaked xml (I needed to add a new element to wrap <Cars>...Net is picky about deserializing arrays):
<?xml version="1.0" encoding="utf-8"?>
<CarCollection>
<Cars>
<Car>
<StockNumber>1020</StockNumber>
<Make>Nissan</Make>
<Model>Sentra</Model>
</Car>
<Car>
<StockNumber>1010</StockNumber>
<Make>Toyota</Make>
<Model>Corolla</Model>
</Car>
<Car>
<StockNumber>1111</StockNumber>
<Make>Honda</Make>
<Model>Accord</Model>
</Car>
</Cars>
</CarCollection>
You have two possibilities.
Method 1. XSD tool
Suppose that you have your XML file in this location C:\path\to\xml\file.xml
Open Developer Command Prompt
You can find it in Start Menu > Programs > Microsoft Visual Studio 2012 > Visual Studio Tools
Or if you have Windows 8 can just start typing Developer Command Prompt in Start screen
Change location to your XML file directory by typing cd /D "C:\path\to\xml"
Create XSD file from your xml file by typing xsd file.xml
Create C# classes by typing xsd /c file.xsd
And that's it! You have generated C# classes from xml file in C:\path\to\xml\file.cs
Method 2 - Paste special
Required Visual Studio 2012+
Copy content of your XML file to clipboard
Add to your solution new, empty class file (Shift+Alt+C)
Open that file and in menu click Edit > Paste special > Paste XML As Classes
And that's it!
Usage
Usage is very simple with this helper class:
using System;
using System.IO;
using System.Web.Script.Serialization; // Add reference: System.Web.Extensions
using System.Xml;
using System.Xml.Serialization;
namespace Helpers
{
internal static class ParseHelpers
{
private static JavaScriptSerializer json;
private static JavaScriptSerializer JSON { get { return json ?? (json = new JavaScriptSerializer()); } }
public static Stream ToStream(this string #this)
{
var stream = new MemoryStream();
var writer = new StreamWriter(stream);
writer.Write(#this);
writer.Flush();
stream.Position = 0;
return stream;
}
public static T ParseXML<T>(this string #this) where T : class
{
var reader = XmlReader.Create(#this.Trim().ToStream(), new XmlReaderSettings() { ConformanceLevel = ConformanceLevel.Document });
return new XmlSerializer(typeof(T)).Deserialize(reader) as T;
}
public static T ParseJSON<T>(this string #this) where T : class
{
return JSON.Deserialize<T>(#this.Trim());
}
}
}
All you have to do now, is:
public class JSONRoot
{
public catalog catalog { get; set; }
}
// ...
string xml = File.ReadAllText(#"D:\file.xml");
var catalog1 = xml.ParseXML<catalog>();
string json = File.ReadAllText(#"D:\file.json");
var catalog2 = json.ParseJSON<JSONRoot>();
The following snippet should do the trick (and you can ignore most of the serialization attributes):
public class Car
{
public string StockNumber { get; set; }
public string Make { get; set; }
public string Model { get; set; }
}
[XmlRootAttribute("Cars")]
public class CarCollection
{
[XmlElement("Car")]
public Car[] Cars { get; set; }
}
...
using (TextReader reader = new StreamReader(path))
{
XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
return (CarCollection) serializer.Deserialize(reader);
}
See if this helps:
[Serializable()]
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlArrayItem(typeof(Car))]
public Car[] Car { get; set; }
}
.
[Serializable()]
public class Car
{
[System.Xml.Serialization.XmlElement()]
public string StockNumber{ get; set; }
[System.Xml.Serialization.XmlElement()]
public string Make{ get; set; }
[System.Xml.Serialization.XmlElement()]
public string Model{ get; set; }
}
And failing that use the xsd.exe program that comes with visual studio to create a schema document based on that xml file, and then use it again to create a class based on the schema document.
I don't think .net is 'picky about deserializing arrays'. The first xml document is not well formed.
There is no root element, although it looks like there is. The canonical xml document has a root and at least 1 element (if at all). In your example:
<Root> <-- well, the root
<Cars> <-- an element (not a root), it being an array
<Car> <-- an element, it being an array item
...
</Car>
</Cars>
</Root>
try this block of code if your .xml file has been generated somewhere in disk and if you have used List<T>:
//deserialization
XmlSerializer xmlser = new XmlSerializer(typeof(List<Item>));
StreamReader srdr = new StreamReader(#"C:\serialize.xml");
List<Item> p = (List<Item>)xmlser.Deserialize(srdr);
srdr.Close();`
Note: C:\serialize.xml is my .xml file's path. You can change it for your needs.
For Beginners
I found the answers here to be very helpful, that said I still struggled (just a bit) to get this working. So, in case it helps someone I'll spell out the working solution:
XML from Original Question. The xml is in a file Class1.xml, a path to this file is used in the code to locate this xml file.
I used the answer by #erymski to get this working, so created a file called Car.cs and added the following:
using System.Xml.Serialization; // Added
public class Car
{
public string StockNumber { get; set; }
public string Make { get; set; }
public string Model { get; set; }
}
[XmlRootAttribute("Cars")]
public class CarCollection
{
[XmlElement("Car")]
public Car[] Cars { get; set; }
}
The other bit of code provided by #erymski ...
using (TextReader reader = new StreamReader(path))
{
XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
return (CarCollection) serializer.Deserialize(reader);
}
... goes into your main program (Program.cs), in static CarCollection XCar() like this:
using System;
using System.IO;
using System.Xml.Serialization;
namespace ConsoleApp2
{
class Program
{
public static void Main()
{
var c = new CarCollection();
c = XCar();
foreach (var k in c.Cars)
{
Console.WriteLine(k.Make + " " + k.Model + " " + k.StockNumber);
}
c = null;
Console.ReadLine();
}
static CarCollection XCar()
{
using (TextReader reader = new StreamReader(#"C:\Users\SlowLearner\source\repos\ConsoleApp2\ConsoleApp2\Class1.xml"))
{
XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
return (CarCollection)serializer.Deserialize(reader);
}
}
}
}
Hope it helps :-)
Kevin's anser is good, aside from the fact, that in the real world, you are often not able to alter the original XML to suit your needs.
There's a simple solution for the original XML, too:
[XmlRoot("Cars")]
public class XmlData
{
[XmlElement("Car")]
public List<Car> Cars{ get; set; }
}
public class Car
{
public string StockNumber { get; set; }
public string Make { get; set; }
public string Model { get; set; }
}
And then you can simply call:
var ser = new XmlSerializer(typeof(XmlData));
var data = (XmlData)ser.Deserialize(XmlReader.Create(PathToCarsXml));
One liner:
var object = (Cars)new XmlSerializer(typeof(Cars)).Deserialize(new StringReader(xmlString));
Try this Generic Class For Xml Serialization & Deserialization.
public class SerializeConfig<T> where T : class
{
public static void Serialize(string path, T type)
{
var serializer = new XmlSerializer(type.GetType());
using (var writer = new FileStream(path, FileMode.Create))
{
serializer.Serialize(writer, type);
}
}
public static T DeSerialize(string path)
{
T type;
var serializer = new XmlSerializer(typeof(T));
using (var reader = XmlReader.Create(path))
{
type = serializer.Deserialize(reader) as T;
}
return type;
}
}
How about a generic class to deserialize an XML document
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
// Generic class to load any xml into a class
// used like this ...
// YourClassTypeHere InfoList = LoadXMLFileIntoClass<YourClassTypeHere>(xmlFile);
using System.IO;
using System.Xml.Serialization;
public static T LoadXMLFileIntoClass<T>(string xmlFile)
{
T returnThis;
XmlSerializer serializer = new XmlSerializer(typeof(T));
if (!FileAndIO.FileExists(xmlFile))
{
Console.WriteLine("FileDoesNotExistError {0}", xmlFile);
}
returnThis = (T)serializer.Deserialize(new StreamReader(xmlFile));
return (T)returnThis;
}
This part may, or may not be necessary. Open the XML document in Visual Studio, right click on the XML, choose properties. Then choose your schema file.
The idea is to have all level being handled for deserialization
Please see a sample solution that solved my similar issue
<?xml version="1.0" ?>
<TRANSACTION_RESPONSE>
<TRANSACTION>
<TRANSACTION_ID>25429</TRANSACTION_ID>
<MERCHANT_ACC_NO>02700701354375000964</MERCHANT_ACC_NO>
<TXN_STATUS>F</TXN_STATUS>
<TXN_SIGNATURE>a16af68d4c3e2280e44bd7c2c23f2af6cb1f0e5a28c266ea741608e72b1a5e4224da5b975909cc43c53b6c0f7f1bbf0820269caa3e350dd1812484edc499b279</TXN_SIGNATURE>
<TXN_SIGNATURE2>B1684258EA112C8B5BA51F73CDA9864D1BB98E04F5A78B67A3E539BEF96CCF4D16CFF6B9E04818B50E855E0783BB075309D112CA596BDC49F9738C4BF3AA1FB4</TXN_SIGNATURE2>
<TRAN_DATE>29-09-2015 07:36:59</TRAN_DATE>
<MERCHANT_TRANID>150929093703RUDZMX4</MERCHANT_TRANID>
<RESPONSE_CODE>9967</RESPONSE_CODE>
<RESPONSE_DESC>Bank rejected transaction!</RESPONSE_DESC>
<CUSTOMER_ID>RUDZMX</CUSTOMER_ID>
<AUTH_ID />
<AUTH_DATE />
<CAPTURE_DATE />
<SALES_DATE />
<VOID_REV_DATE />
<REFUND_DATE />
<REFUND_AMOUNT>0.00</REFUND_AMOUNT>
</TRANSACTION>
</TRANSACTION_RESPONSE>
The above XML is handled in two level
[XmlType("TRANSACTION_RESPONSE")]
public class TransactionResponse
{
[XmlElement("TRANSACTION")]
public BankQueryResponse Response { get; set; }
}
The Inner level
public class BankQueryResponse
{
[XmlElement("TRANSACTION_ID")]
public string TransactionId { get; set; }
[XmlElement("MERCHANT_ACC_NO")]
public string MerchantAccNo { get; set; }
[XmlElement("TXN_SIGNATURE")]
public string TxnSignature { get; set; }
[XmlElement("TRAN_DATE")]
public DateTime TranDate { get; set; }
[XmlElement("TXN_STATUS")]
public string TxnStatus { get; set; }
[XmlElement("REFUND_DATE")]
public DateTime RefundDate { get; set; }
[XmlElement("RESPONSE_CODE")]
public string ResponseCode { get; set; }
[XmlElement("RESPONSE_DESC")]
public string ResponseDesc { get; set; }
[XmlAttribute("MERCHANT_TRANID")]
public string MerchantTranId { get; set; }
}
Same Way you need multiple level with car as array
Check this example for multilevel deserialization
If you're getting errors using xsd.exe to create your xsd file, then use the XmlSchemaInference class as mentioned on msdn. Here's a unit test to demonstrate:
using System.Xml;
using System.Xml.Schema;
[TestMethod]
public void GenerateXsdFromXmlTest()
{
string folder = #"C:\mydir\mydata\xmlToCSharp";
XmlReader reader = XmlReader.Create(folder + "\some_xml.xml");
XmlSchemaSet schemaSet = new XmlSchemaSet();
XmlSchemaInference schema = new XmlSchemaInference();
schemaSet = schema.InferSchema(reader);
foreach (XmlSchema s in schemaSet.Schemas())
{
XmlWriter xsdFile = new XmlTextWriter(folder + "\some_xsd.xsd", System.Text.Encoding.UTF8);
s.Write(xsdFile);
xsdFile.Close();
}
}
// now from the visual studio command line type: xsd some_xsd.xsd /classes
You can just change one attribute for you Cars car property from XmlArrayItem to XmlElment. That is, from
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlArrayItem(typeof(Car))]
public Car[] Car { get; set; }
}
to
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlElement("Car")]
public Car[] Car { get; set; }
}
My solution:
Use Edit > Past Special > Paste XML As Classes to get the class in your code
Try something like this: create a list of that class (List<class1>), then use the XmlSerializer to serialize that list to a xml file.
Now you just replace the body of that file with your data and try to deserialize it.
Code:
StreamReader sr = new StreamReader(#"C:\Users\duongngh\Desktop\Newfolder\abc.txt");
XmlSerializer xml = new XmlSerializer(typeof(Class1[]));
var a = xml.Deserialize(sr);
sr.Close();
NOTE: you must pay attention to the root name, don't change it. Mine is "ArrayOfClass1"
I have small problem - XML deserialization completely ignores items, which are out of alphabetic order. In example object (description in end of question), Birthday node is after FirstName node, and it is ignored and assigned default value after deserialization. Same for any other types and names (I had node CaseId of Guid type after node Patient of PatientInfo type, and after deserialization it had default value).
I'm serializing it in one application, using next code:
public static string SerializeToString(object data)
{
if (data == null) return null;
XmlSerializerNamespaces ns = new XmlSerializerNamespaces();
ns.Add("", "");
// what should the XmlWriter do?
var settings = new XmlWriterSettings
{
OmitXmlDeclaration = true,
NewLineChars = ""
};
using (var stringwriter = new System.IO.StringWriter())
{
// Use an XmlWriter to wrap the StringWriter
using (var xmlWriter = XmlWriter.Create(stringwriter, settings))
{
var serializer = new XmlSerializer(data.GetType(), "");
// serialize to the XmlWriter instance
serializer.Serialize(xmlWriter, data, ns);
return stringwriter.ToString();
}
}
}
Such approach was used to get proper result as argument for WebMethod (full problem described here). Results are something like this:
<PatientInfo><FirstName>Foo</FirstName><Birthday>2015-12-19T16:21:48.4009949+01:00</Birthday><RequestedClientID>00000000-0000-0000-0000-000000000000</RequestedClientID>00000000-0000-0000-0000-000000000000</patientId></PatientInfo>
Also I'm deserializing it in another application in simple manner
public static T Deserialize<T>(string xmlText)
{
if (String.IsNullOrEmpty(xmlText)) return default(T);
using (var stringReader = new StringReader(xmlText))
{
var serializer = new XmlSerializer(typeof(T));
return (T)serializer.Deserialize(stringReader);
}
}
Example object:
[XmlRoot("PatientInfo")]
public class PatientInfo
{
[XmlElement("FirstName")]
public string FirstName { get; set; }
[XmlElement("LastName")]
public string LastName { get; set; }
[XmlElement("SSN")]
public string SSN { get; set; }
[XmlElement("Birthday")]
public DateTime? Birthday { get; set; }
[XmlElement("RequestedClientID")]
public Guid RequestedClientID { get; set; }
[XmlElement("patientId")]
public Guid patientId { get; set; }
}
So, I'd like to have answer for one of two questions - 1) How can I adjust my serialization to have all items in alphabetical order? 2) How can I adjust my deserialization, so it won't ignore items out of alphabetical order?
Any help is appreciated.
Update:
Just figured out, that deserialization method I'm using is not actually used at all in my problem, since I'm using serialized info as data with WebMethod, and it is deserialized with some internal mechanism of WCF.
WCF uses DataContractSerializer. This serializer is sensitive to XML element order, see Data Member Order. There's no quick way to disable this, instead you need to replace the serializer with XmlSerializer.
To do this, see Using the XmlSerializer Class, then and apply [XmlSerializerFormat] to your service, for instance:
[ServiceContract]
[XmlSerializerFormat]
public interface IPatientInfoService
{
[OperationContract]
public void ProcessPatientInfo(PatientInfo patient)
{
// Code not shown.
}
}
[XmlRoot("PatientInfo")]
public class PatientInfo
{
[XmlElement("FirstName")]
public string FirstName { get; set; }
[XmlElement("LastName")]
public string LastName { get; set; }
[XmlElement("SSN")]
public string SSN { get; set; }
[XmlElement("Birthday")]
public DateTime? Birthday { get; set; }
[XmlElement("RequestedClientID")]
public Guid RequestedClientID { get; set; }
[XmlElement("patientId")]
public Guid patientId { get; set; }
}
I have a struct like this:
public struct Vehicles
{
public string Name { get; set; }
public string Count { get; set; }
public List<Car> Cars { get; set; }
}
public struct Car
{
public string Name { get; set; }
public int Count { get; set; }
public List<Tire> Tires { get; set; }
}
public struct Tire
{
public string Brand { get; set; }
public int Count { get; set; }
public int UniqueCount { get; set; }
public List<Dimension> Dimensions { get; set; }
}
public struct Dimension
{
public string Size { get; set; }
public int AlternateSize { get; set; }
}
When I serialize "Vehicles" it is like:
<root xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xsd="http://www.w3.org /2001/XMLSchema">
<Vehicles>
<Name>SuperVehicles</Name>
<Cars>
<Car>
<Name>BMW</Name>
<Count>29</Count>
<Tires>
<Tire>
<Name>DMZ</Name>
<Count>26</Count>
<UniqueCount>24</UniqueCount>
<Dimensions>
<Dimension>
<Size>70x570</Size>
<AlternateSize>70x580</AlternateSize>
</Dimension>
<Dimension>
<Size>60x570</Size>
<AlternateSize>60x580</AlternateSize>
</Dimension>
<Dimension>
<Size>50x570</Size>
<AlternateSize>50x580</AlternateSize>
</Dimension>
</Dimensions>
</Tire>
</Tires>
</Car>
</Cars>
</Vehicles>
</root>
Now the problem is, I want to serialize it like this:
<root>
<vehicles vehicleName="superVehicles" vehicleCount="50" carName="BMW"
carCount="25" tireBrand="kamu" tireCount="15" tireUniqueCount="15"
dimensionSize="70x570" dimensionAlternateSize="70x580" />
<vehicles vehicleName="superVehicles" vehicleCount="35" carName="MERCEDES"
carCount="22" tireBrand="kamu" tireCount="12" tireUniqueCount="12"
dimensionSize="60x570" dimensionAlternateSize="60x580" />
<vehicles vehicleName="superVehicles" vehicleCount="35" carName="PORSCHE"
carCount="22" tireBrand="kamu" tireCount="12" tireUniqueCount="12"
dimensionSize="60x570" dimensionAlternateSize="60x580" />
</root>
Do I have to change the structure and avoid the groupings or is there any way to create a schema for xml serialization to gather this result.
Summary:
I get all the child items in a new tag when I serialize the root struct to xml but I need to take them as properties of an instance that create only the count of root (Vehicles in this situation) element of rows to xml.
You need to do manual serialization.
Here is how you can implement this using System.Xml.Linq :
var xmlElementsVehicles = new[]{
new XElement("vehicles ", new object[]
{
new XElement("vehicleName", "superVehicles"),
new XElement("vehicleCount", 35),
new XElement("carName", "PORSCHE"),
new XElement("carCount", 2)
}),
new XElement("vehicles ", new object[]
{
new XElement("vehicleName", "superVehicles"),
new XElement("vehicleCount", 35),
new XElement("carName", "PORSCHE"),
new XElement("carCount", 2)
})
};
var root = new XElement("root", xmlElementsVehicles );
var myXml = new XDocument(new XDeclaration("1.0", "utf-8", "yes"), root);
using (var xmlWriter = XmlWriter.Create(stream))
{
myXml.Save(xmlWriter);
}
To use XmlSerializer the model must roughly be the same as the layout; a few things can change (names, etc). However, your model is nothing like the XML. Three options, then:
create a second DTO model that looks like the XML (you can use xsd.exe on the sample XML to automate this), and use XmlSerializer
don't use XmlSerializer, but build the XML somehow else (XmlDocument or XDocument would be the obvious two, or XmlWriter if the size is very large)
use something like xslt to reshape the XML after writing
There is nothing "easy" can be done to make XmlSerializer write that model into your desired XML.
I'm converting a Dictionary object into a List<> derived class by means of the following two declarations:
[Serializable]
public class LogItem
{
public string Name { get; set; }
public string Value { get; set; }
public LogItem(string key, string value)
{
Name = key; Value = value;
}
public LogItem() { }
}
public class SerializableDictionary : List<LogItem>
{
public SerializableDictionary(Dictionary<string, string> table)
{
IDictionaryEnumerator index = table.GetEnumerator();
while (index.MoveNext())
{
Put(index.Key.ToString(), index.Value.ToString());
}
}
private void Put(string key, string value)
{
base.Add(new LogItem(key, value));
}
}
I intend to serialize SerializableDictionary by means of the following code:
SerializableDictionary log = new SerializableDictionary(contents);
using (StringWriter xmlText = new StringWriter())
{
XmlSerializer xmlFormat =
new XmlSerializer(typeof(SerializableDictionary), new XmlRootAttribute("Log"));
xmlFormat.Serialize(xmlText, log);
}
Works fine, but I'm unable to change the XML formatting.
This XML document is intended to be sent to a xml database field and is not meant to be deserialized.
I'd rather have both Name and Value
formatted as attributes of the
LogItem element.
However any attempts at using XMLAttribute have resulted in Reflection or compilation errors. I'm at loss as to what I can do to achieve this requirement. Could someone please help?
you can annotate them with XmlAttribute:
[Serializable]
public class LogItem
{
[XmlAttribute]
public string Name { get; set; }
[XmlAttribute]
public string Value { get; set; }
public LogItem(string key, string value)
{
Name = key; Value = value;
}
public LogItem() { }
}
This worked fine for me and produced the following XML (based on sample input):
<?xml version="1.0" encoding="utf-16"?>
<Log xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<LogItem Name="foo" Value="bar" />
<LogItem Name="asdf" Value="bcxcvxc" />
</Log>
There must be something else going on besides what you are showing. I am able to compile and execute the above code both with and without the XmlAttribute attribute applied to the Name and Value properties.
[XmlAttribute]
public string Name { get; set; }
[XmlAttribute]
public string Value { get; set; }
I'm trying to get to this result while serializing XML:
<Root Name="blah">
<SomeKey>Eldad</SomeKey>
<Element>1</Element>
<Element>2</Element>
<Element>3</Element>
<Element>4</Element>
</root>
Or in other words - I'm trying to contain an array within the "root" element, alongside additional keys.
This is my crude attempt:
[XmlRootAttribute(ElementName="Root", IsNullable=false)]
public class RootNode
{
[XmlAttribute("Name")]
public string Name { get; set; }
public string SomeKey { get; set; }
[XmlArrayItem("Element")]
public List<int> Elements { get; set; }
}
And my serialization:
string result;
XmlSerializer serializer = new XmlSerializer(root.GetType());
StringBuilder sb = new StringBuilder();
using (StringWriter sw = new StringWriter(sb))
{
serializer.Serialize(sw, root);
result = sw.ToString();
}
However, this is my result (Removed the namespace for clarity):
<Root>
<SomeKey>Eldad</SomeKey>
<Elements>
<Element>1</Element>
<Element>2</Element>
<Element>3</Element>
</Elements>
</Root>
Is there any way to remove the "Elements" part?
Use XmlElement attribute on the Array, this will instruct the serializer to serialize the array items as child elements of the current element and not create a new root element for the array.
[XmlRootAttribute(ElementName="Root", IsNullable=false)]
public class RootNode
{
[XmlAttribute("Name")]
public string Name { get; set; }
public string SomeKey { get; set; }
[XmlElement("Element")]
public List<int> Elements { get; set; }
}
Thanks to Chris Taylor for the answer to my problem too. Using an asmx web service, I was getting the following XML:
<Manufacturers>
<Manufacturer>
<string>Bosch</string>
<string>Siemens</string>
</Manufacturer>
</Manufacturers>
I wanted to get the manufacturer names directly in the the element, getting rid of the element like this:
<Manufacturers>
<Manufacturer>Bosch</Manufacturer>
<Manufacturer>Siemens</Manufacturer>
</Manufacturers>
For anyone else with the same problem, my code to achieve this (in VB.Net) is:
<WebMethod()> _
Public Function GetManufacturers() As Manufacturers
Dim result As New Manufacturers
result.Manufacturer.Add("Bosch")
result.Manufacturer.Add("Siemens")
Return result
End Function
<XmlRoot(ElementName:="Manufacturers")> _
Public Class Manufacturers
<XmlElement("Manufacturer")> _
Public Manufacturer As New List(Of String)
End Class