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 tried to Deserialize an XML to an object using C#, but I am getting some errors.
My XML file is:
<?xml version="1.0" encoding="UTF-8"?>
<ns0:transmission xmlns:ns0="blabla.xsd">
<ns0:DiagnosisErrorResponse>
<ns0:ID>7</ns0:ID>
<ns0:ErrorCode>9</ns0:ErrorCode>
<ns0:ErrorDescription>sometext</ns0:ErrorDescription>
<ns0:ErrorDate>11-12-2018</ns0:ErrorDate>
</ns0:DiagnosisErrorResponse>
</ns0:transmission>
I have an entity like this:
[XmlRoot(Namespace = "blabla.xsd", ElementName = "ns0:transmission", DataType = "string", IsNullable = true)]
public class DiagnosisErrorResponse
{
[XmlElement("ID")]
public long ID { get; set; }
[XmlElement("ErrorCode")]
public int ErrorCode { get; set; }
[XmlElement("ErrorDescription")]
public string ErrorDescription { get; set; }
[XmlElement("ErrorDate")]
public string ErrorDate { get; set; }
}
and my error is:
<transmission xmlns='blabla.xsd'> was not expected.
My function is this:
private void ReadXmlFileByPath(string filePath)
{
string xmlText = string.Empty;
XmlDocument d = new XmlDocument();
using (XmlTextReader tr = new XmlTextReader(filePath))
{
tr.Namespaces = true;
d.Load(tr);
xmlText = d.InnerXml;
}
XmlSerializer xs = new XmlSerializer(typeof(DiagnosisErrorResponse), "");
using (TextReader reader = new StringReader(xmlText))
{
DiagnosisErrorResponse result = (DiagnosisErrorResponse)xs.Deserialize(reader);
}
}
Update: I removed the 'ns0:' prefix from the XmlElement attribute at entity but I still get the same error.
Your problem is that your c# data model does not match your XML. Specifically:
You have indicated that the element name for DiagnosisErrorResponse should be "ns0:transmission". The namespace prefix ns0: should not be included, it is just a lookup in the XML file to find the actual namespace, which is defined by the xmlns attribute xmlns:ns0="blabla.xsd".
The XML has a level of nesting not accounted for in your data model. The XML in your question is not indented; if I indent it, it looks like:
<ns0:transmission xmlns:ns0="blabla.xsd">
<ns0:DiagnosisErrorResponse>
<ns0:ID>7</ns0:ID>
<ns0:ErrorCode>9</ns0:ErrorCode>
<ns0:ErrorDescription>sometext</ns0:ErrorDescription>
<ns0:ErrorDate>11-12-2018</ns0:ErrorDate>
</ns0:DiagnosisErrorResponse>
</ns0:transmission>
As can now be seen, ID and so on are not children of transmission, they are children of DiagnosisErrorResponse which is a child of transmission. This needs to be accounted for in your c# classes.
Combining these two issues, your data model should look like:
[XmlRoot(Namespace = "blabla.xsd", ElementName = "transmission", IsNullable = true)]
public class Transmission
{
[XmlElement("DiagnosisErrorResponse")]
public DiagnosisErrorResponse DiagnosisErrorResponse { get; set; }
}
[XmlRoot(Namespace = "blabla.xsd", IsNullable = true)]
public class DiagnosisErrorResponse
{
[XmlElement("ID")]
public long ID { get; set; }
[XmlElement("ErrorCode")]
public int ErrorCode { get; set; }
[XmlElement("ErrorDescription")]
public string ErrorDescription { get; set; }
[XmlElement("ErrorDate")]
public string ErrorDate { get; set; }
}
And your deserialization code should look like:
private DiagnosisErrorResponse ReadXmlFileByPath(string filePath)
{
using (var xmlReader = XmlReader.Create(filePath))
{
XmlSerializer xs = new XmlSerializer(typeof(Transmission));
var result = (Transmission)xs.Deserialize(xmlReader);
return result.DiagnosisErrorResponse;
}
}
Demo fiddle #1 here.
Alternatively, if you were trying to deserialize just the <DiagnosisErrorResponse> element inside the XML file without creating a data model for the root element, you could use XmlReader to scan forward in the XML until the appropriate element is found:
private DiagnosisErrorResponse ReadXmlFileByPath(string filePath)
{
using (var xmlReader = XmlReader.Create(filePath))
{
if (!xmlReader.ReadToDescendant("DiagnosisErrorResponse", "blabla.xsd"))
return null;
XmlSerializer xs = new XmlSerializer(typeof(DiagnosisErrorResponse));
var result = (DiagnosisErrorResponse)xs.Deserialize(xmlReader);
return result;
}
}
Demo fiddle #2 here.
Notes:
There is no need to load the XML file into an XmlDocument, re-serialize it as a string, then deserialize the string. The XML can be deserialized directly from a file in one step using an XmlReader or StreamReader.
XmlTextReader has been deprecated by Microsoft in its documentation. XmlReader.Create() should be used instead.
You can avoid modeling errors such as the ones above by using one of the code generation tools mentioned in Generate C# class from XML.
The easiest way to diagnose an error in XML deserialization is to serialize an instance of your root type, then compare the resulting XML with the XML to be deserialized. Inconsistencies are typically easily spotted and indicate where you have problems.
I'm trying to deserialize a simple XML, but it Always returns null to my object.
XML
<ns0:Customer xmlns:ns0="http://myNameSpace.Customer">
<Company>001</Company>
<Division>003</Division>
</ns0:Customer>
CLASS
[XmlRoot(ElementName = "Customer", Namespace = "http://myNameSpace.Customer")]
public class Customer
{
[XmlElement(ElementName = "Company")]
public string Company { get; set; }
[XmlElement(ElementName = "Division")]
public string Division { get; set; }
}
CODE
File.AppendAllText(fileName, string.Format("Polling at {0}\n", DateTime.Now.ToString("o")));
XmlSerializer serializer = new XmlSerializer(typeof(Customer));
using (FileStream fileStream = new FileStream(PlatformServices.Default.Application.ApplicationBasePath
+ #"\Customers\Customer.xml", FileMode.Open))
{
var customers = ((Customer)serializer.Deserialize(fileStream));
}
RESULT
Any ideas ?
Your containing element does not define a default namespace, so the default namespace of the document is an empty string.
This means that <Company> and <Division> don't have the same namespace as the containing element, and inherit the default namespace "".
Your example will work if you rewrite the attributes to take this into account:
[XmlRoot(ElementName = "Customer", Namespace = "http://myNameSpace.Customer")]
public class Customer
{
[XmlElement(ElementName = "Company", Namespace = "")]
public string Company { get; set; }
[XmlElement(ElementName = "Division", Namespace = "")]
public string Division { get; set; }
}
as is demonstrated below:
var xmlStr = #" <ns0:Customer xmlns:ns0=""http://myNameSpace.Customer"">
<Company>001</Company>
<Division>003</Division>
</ns0:Customer>";
var ms = new MemoryStream(Encoding.UTF8.GetBytes(xmlStr));
XmlSerializer serializer = new XmlSerializer(typeof(Customer));
var customer = ((Customer)serializer.Deserialize(ms));
//yay. fully populated
(First, feel free to edit my title, I really don't find a better one for my problem)
I got my root class:
[XmlRoot("ProductList")]
public class Product
{
[XmlElement("Property1")]
public string property1 { get; set; }
[XmlElement("Property2")]
public Property2 property2 { get; set; }
[XmlArray("Property3Array")]
[XmlArrayItem("Property3ArrayItem")]
public List<Property3> property3{ get; set; }
}
I'm serializing the List of Products like this:
public void Execute(IJobExecutionContext context)
{
var products = _productionService.GetAllProducts();
XmlSerializer xs = new XmlSerializer(typeof(List<Product>));
using (StreamWriter sw = new StreamWriter("products.xml"))
{
xs.Serialize(sw, products);
}
}
The serialization is working properly BUT, in my products.xml file, the root node is:
<ArrayOfProduct xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
...
</ArrayOfProduct>
But I want the root element of the list to be named ProductList
I tried with [XmlRoot("ProductList")] But that doesn't work.
So, how can I rename the name of the root xml object of a List<Class> ?
Use a constructor overload that accepts an XmlRootAttribute.
var xs = new XmlSerializer(typeof(List<Product>), new XmlRootAttribute("ProductList"));
See this fiddle for a working demo.
I wanna get xml file from http and convert it to object.
So right now I have 2 methods: one to get http response body like that:
var httpClient = new HttpClient();
var op = httpClient.GetStringAsync(uri);
var httpResponseBody = "";
try {
var httpResponse = await httpClient.GetAsync(uri);
httpResponse.EnsureSuccessStatusCode();
httpResponseBody = await httpResponse.Content.ReadAsStringAsync();
return httpResponseBody;
}
...
which returns string httpResponseBody.
Second one tries to convert this xml in string to object:
res = await task;
var reader = new XmlSerializer(typeof(Schedule));
using (var tr = new MemoryStream(Encoding.UTF8.GetBytes(res)))
{
var schedule = (Schedule)reader.Deserialize(tr);
return schedule;
}
The problem is that the content I receive is in different encoding and I don't know how to convert it to make deserialization possible.
I am getting something like this:
<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<ramowka><dzien name=\"PoniedziaÅ\u0082ek\" count=\"2\"/></ramowka>\n
How to get rid of '\n' and Å\u0082 (should be ł) ?
Right now I am getting Exception from reader.Deserialize: {"<ramowka xmlns=''> was not expected."}
Schedule class:
[XmlType(AnonymousType = true)]
[XmlRootAttribute(Namespace = "", IsNullable = false)]
public class Schedule
{
[XmlElementAttribute("ramowka")]
public ScheduleDay[] AuditionDays { get; set; }
}
I've changed Schedule class to:
[XmlType(AnonymousType = true)]
[XmlRootAttribute("ramowka")]
public class Schedule
{
[XmlElementAttribute("dzien")]
public ScheduleDay[] AuditionDays { get; set; }
}
Now it looks like working. Thanks Petter for hint with Root attribute.
Setting the root object on the XmlSerializer fixes the problem:
var reader = new XmlSerializer(typeof(Schedule), new XmlRootAttribute("ramowka"));
...though I used slightly different attributes:
[DataContract]
public class ScheduleDay
{
[DataMember, XmlAttribute]
public string name { get; set; }
[DataMember, XmlAttribute]
public string count { get; set; }
}
[DataContract]
public class Schedule
{
[DataMember]
public ScheduleDay dzien { get; set; }
}
I haven't tried yours yet, but these work.
For a collection of ScheduleDays, this combo works:
[XmlType("dzien")]
public class ScheduleDay
{
[XmlAttribute]
public string name { get; set; }
[XmlAttribute]
public string count { get; set; }
}
Usage:
XmlSerializer reader = new XmlSerializer(typeof(List<ScheduleDay>), new XmlRootAttribute("ramowka"));
using (MemoryStream stream = new MemoryStream(Encoding.UTF8.GetBytes(Xml)))
{
List<ScheduleDay> schedule = (List<ScheduleDay>)reader.Deserialize(stream);
}
The Schedule class just disappeared from the equation.
Escapes in the HTML
The \ns are part of the XML structure, so no need to worry about those. The deserializer will translate \u0082 into its equivalent character, which is
BREAK PERMITTED HERE. Which you probably don't want. The Å looks out of place too -- it's the last letter of the Norwegian alphabet and not used in Polish, AFAIK.