Question

    Public Class Employee
{
   Public String EmployeeId {get;set;}
   Public String EmployeeName {get;set;}
   Public String Department {get;set;}
}

Public Class Department
{
   Public String DepartmentId {get;set;}
   Public String DepartmentName {get;set;}
   Public String Address {get;set;}
}

Public Class Address
{
   Public String AddrOne {get;set;}
   Public String City {get;set;}
}

我有3个模型，列表，列表和列表执行程序后，上面提到的3个模型应该填充List，List和List 我必须以下面的格式返回数据......

以下格式获得回复的最佳方法是什么？

<Employees>
 <Employee>
   <EmployeeID>   </EmployeeID>
   <EmployeeName>   </EmployeeName>
   <Department>
     <DepartmentID>     </DepartmentID>
     <DepartmentName>     </DepartmentName>
    <Address>
      <Addr1>   </Addr1>
          <City>    </City>
    </Address>
   <Department>
 </Employee>
</Employees>

Answer 1

您可以创建以下类：

public class Model
{
   public List<Employee> Employees { get; set; }
}

public class Employee
{
   public string EmployeeId { get; set; }
   public string EmployeeName { get; set; }
   public Department Department { get; set; }
}

public class Department
{
   public string DepartmentId { get; set; }
   public string DepartmentName {get; set; }
   public Address Address { get; set; }
}

public class Address
{
   public string AddrOne { get; set; }
   public string City { get; set; }
}

接下来，您可以创建模型的实例并填充数据并将其序列化为XML

Answer 2

XML序列化：

创建一个CollectionClass并添加方法来序列化它：

public class MyCollection
{
    public List<Employee> = new List<Employee>();
    public List<Department> = new List<Department>();
    public List<Address> = new List<Address>();

    public string ToXML()
    {
        var stringwriter = new System.IO.StringWriter();
        var serializer = new XmlSerializer(this.GetType());
        serializer.Serialize(stringwriter, this);
        return stringwriter.ToString();
    }

    // You have to use your Class-Type here 3 times
    public static MyCollection LoadFromXML(string filePath)
    {
        using (StreamReader streamReader = new System.IO.StreamReader(filePath))
        {
            var serializer = new XmlSerializer(typeof(MyCollection));
            return serializer.Deserialize(streamReader) as MyCollection;
        }
    }
}

现在您可以将您的类保存为xml-File：

        MyCollection myCollection = new MyCollection();
        //Now add your entries, myCollection.Add(new Department(....));  

        //Save your class as xml-File
        File.WriteAllText("C:\\MyClass.xml", myCollection.ToXML());

然后你可以加载它：

        //Load your class
        MyCollection myCollection = MyCollection.LoadFromXML("C:\\MyClass.xml");

编辑：将其更改为CollectionClass-Sample，它应该适合您的情况

如何将List <model>转换为复杂的响应类型？

2 个答案: