How to Deserialize XML document

My solution:

1. Use `Edit > Past Special > Paste XML As Classes` to get the class in your code
2. Try something like this: create a list of that class (`List<class1`>), then use the `XmlSerializer` to serialize that list to a `xml` file.
3. 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"

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.

# 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 :-)

How about you just save the xml to a file, and use [xsd][1] to generate C# classes?

1. Write the file to disk (I named it foo.xml)
2. Generate the xsd: `xsd foo.xml`
3. 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)


[1]:

[To see links please register here]

One liner:
```
var object = (Cars)new XmlSerializer(typeof(Cars)).Deserialize(new StringReader(xmlString));
```

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][1] 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>


[1]:

[To see links please register here]

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));