如何将 List<Model> 转换为复杂响应类型？答案

【问题标题】：How to convert List<Model> to complex response type?如何将 List<Model> 转换为复杂响应类型？
【发布时间】：2017-02-13 15:44:43
【问题描述】：

    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，List、List 和 List 程序执行后，上述 3 个模型应填充 List、List 和 List 我必须以以下格式返回数据...

获得以下格式响应的最佳方法是什么？

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

【问题讨论】：

XML serialization
@AfnanAhmad ，我正在循环所有 3 个集合（员工、部门和地址），基于它们的层次结构并创建 xml 结构。我相信有更好的方法来做到这一点......但没有任何线索 forrach(DataRow drEmp in ...) { xmlElement.Add(EmpID) xmlElement.Add(EmpNAme) forrach(DataRow drEmp in ...) { xmlElement.Add(DeptID) xmlElement.Add(DeptName) forrach(DataRow drEmp in ...) { xmlElement.Add(Addr1) } } }

标签： c# linq linq-to-xml

【解决方案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，您可以创建模型实例并填充数据并将其序列化为 XML

【讨论】：

【解决方案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;
        }
    }
}

现在您可以将类保存为 xml 文件：

        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());

然后你就可以加载它了：

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

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

【讨论】：