Question

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

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

和简单的方法，返回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"
            }
        };
    }

的Program.cs：

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

我有文件结果：

<?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>

但我想Root Element的名称是：“Employees”，而不是“ArrayOfEmployee” 我该怎么做？

我想这样做，因为我有文件，结构如下：

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

Answer 1

只需更改如下

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

全部。但是要在你的问题中得到一个干净的xml（没有xml声明，没有xsi或xsd命名空间等），你应该使用一些技巧

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);

Answer 2

您可以传递XmlRootAttribute来设置元素名称：

var root = new XmlRootAttribute("Employees");

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

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

... root参数允许您替换默认对象通过指定XmlRootAttribute来获取信息;该对象允许你设置不同的命名空间，元素名称，等等。

Answer 3

您可以使用属性标记属性，使用XmlArray和XmlArrayItem属性

XML Parse列表

3 个答案: