XML Parse to list

Question

[XmlRoot("Employees")]
public class Employee
{
    [XmlElement("EmpId")]
    public int Id { get; set; }

    [XmlElement("Name")]
    public string Name { get; set; }
}

and simple method, which return List:

    public static List<Employee> SampleData()
    {
        return new List<Employee>()
        {
            new Employee(){
                Id   = 1,
                Name = "pierwszy"
            },
            new Employee(){
                Id   = 2,
                Name = "drugi"
            },
            new Employee(){
                Id   = 3,
                Name = "trzeci"
            }
        };
    }

Program.cs:

   var list = Employee.SampleData();
   XmlSerializer ser = new XmlSerializer(typeof(List<Employee>));
   TextWriter writer = new StreamWriter("nowi.xml");
   ser.Serialize(writer, list);

I have file result:

<?xml version="1.0" encoding="utf-8"?>
<ArrayOfEmployee xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Employee>
    <EmpId>1</EmpId>
    <Name>pierwszy</Name>
  </Employee>
  <Employee>
    <EmpId>2</EmpId>
    <Name>drugi</Name>
  </Employee>
  <Employee>
    <EmpId>3</EmpId>
    <Name>trzeci</Name>
  </Employee>
</ArrayOfEmployee>

but i would like for Root Element has name: "Employees", not "ArrayOfEmployee" how can i make it?

I want to do it, because i have file, where structure looks like:

<Employees>
    <Employee>
    ...
    </Employee>
    <Employee>
    ...
    </Employee>
</Employees>

possible duplicate of [How to rename XML attribute that generated after serializing List of objects](http://stackoverflow.com/questions/3129001/how-to-rename-arrayof-xml-attribute-that-generated-after-serializing-list-of-o) — tnw, Aug 07 '14 at 19:30

L.B · Answer 1 · 2014-08-07T20:06:26.770

Just change as below

XmlSerializer ser = new XmlSerializer(typeof(List<Employee>), 
                                      new XmlRootAttribute("Employees"));

that's all. But to get a clean xml as in your question (no xml declaration, no xsi or xsd namespaces etc.), you should use a few tricks

XmlSerializer ser = new XmlSerializer(typeof(List<Employee>), 
                                      new XmlRootAttribute("Employees"));

TextWriter writer = new StreamWriter(filename);
var xmlWriter = XmlWriter.Create(writer, new XmlWriterSettings() { OmitXmlDeclaration = true, Indent = true });

XmlSerializerNamespaces ns = new XmlSerializerNamespaces();
ns.Add("", "");

ser.Serialize(xmlWriter, list, ns);

score 1 · Answer 2 · answered Aug 07 '14 at 19:30

You can pass the XmlRootAttribute to set the element Name:

var root = new XmlRootAttribute("Employees");

XmlSerializer ser = new XmlSerializer(typeof(List<Employee>), root);
TextWriter writer = new StreamWriter("nowi.xml");
ser.Serialize(writer, list);

From http://msdn.microsoft.com/en-us/library/f1wczcys%28v=vs.110%29.aspx :

... the root parameter allows you to replace the default object's information by specifying an XmlRootAttribute; the object allows you to set a different namespace, element name, and so on.

score 0 · Answer 3 · answered Aug 07 '14 at 21:15

0

You can mark your property with attributes, use the XmlArray and XmlArrayItem attribute

answered Aug 07 '14 at 21:15

Amjad Abdelrahman

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XML Parse to list

3 Answers3