如何从Xml文件存储和检索对象

Question

I have this class:

public class MyMenu
{
    public string Name { get; set; }
    public string Type { get; set; }
}

This class I want to use it in a dynamic menu, and I do not want to store his data in a database.

I want to store its data in Xml file.

Till now I have this for saving data:

string path = Server.MapPath("~/Content/Files");
XmlSerializer serial = new XmlSerializer(model.GetType());
System.IO.StreamWriter writer = new System.IO.StreamWriter(path + "\\ribbonmenu");
serial.Serialize(writer, model);
writer.Close();

And this to get the data:

string path = Server.MapPath("~/Content/Files");
XmlSerializer serial = new XmlSerializer(typeof(RibbonMenu));
System.IO.StreamReader reader = new System.IO.StreamReader(path + "\\ribbonmenu");          
RibbonMenu menu =(RibbonMenu) serial.Deserialize(reader);
reader.Close();

What I have is working for one object to save and get. I need to save multiple objects, and get the collection of objects, something like:

IEnumerable<MyMenu> model=(IEnumerable<MyMenu>) serial.Deserialize(reader);

Can someone give me a solution? Thanks.

Edit: The content of the generated Xml with my code is:

<?xml version="1.0" encoding="utf-8"?>
<MyMenu xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Id>0</Id>
  <Menu>Home</Menu>
  <Type>Button</Type>

</MyMenu>

Answer 1

When serializing you should initialize a collection like this:

var model = new List<MyMenu>()
{
    new MyMenu() { Name = "Menu1", Type = "Ribbon" },
    new MyMenu() { Name = "Menu2", Type = "Ribbon" },
};

This way, when you serialize you'd get something like this:

<?xml version="1.0" encoding="utf-8"?>
<ArrayOfMyMenu xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <MyMenu>
    <Name>Menu1</Name>
    <Type>Ribbon</Type>
  </MyMenu>
  <MyMenu>
    <Name>Menu2</Name>
    <Type>Ribbon</Type>
  </MyMenu>
</ArrayOfMyMenu>

And you can get the object back by using List as the type of serializer:

XmlSerializer serial = new XmlSerializer(typeof(List<MyMenu>));
System.IO.StreamReader reader = new System.IO.StreamReader("ribbonmenu.xml");
var menu = (List<MyMenu>)serial.Deserialize(reader);
reader.Close(); 

Hope this helps.