如何将对象集合与复杂对象属性绑定？

Question

我正在尝试加载由 Employee 对象集合组成的 XML 数据。以下函数适用于简单数据类型（如 String 和 Int）的属性。我想知道如何导入复杂类型的数据类型。例如，

这个函数工作正常：

private void LoadData()
{
   XDocument employeesDoc = XDocument.Load("Employees.xml");
   List<Employee> data = (from employee in employeesDoc.Descendants("Employee")
      select new Employee
      {
         FirstName= employee.Attribute("FirstName").Value,
         LastName = employee.Attribute("LastName ").Value,
         PhoneNumber = employee.Attribute("PhoneNumber").Value
      }).ToList();
  Employees.ItemsSource = data;
}

这是 Employee 类：

public class Employee
{
  public int Id { get; set; }
  public string FirstName { get; set; }
  public string LastName { get; set; }
  public string PhoneNumber { get; set; }
  public Department Department { get; set; }
}

这是 Department 类：

public class Department
{
  public int Id { get; set; }
  public string Name { get; set; }
  public string Description { get; set; }
  public Employee Manager { get; set; }
}

那么，如果我的 XML 文件看起来像这样：

<Employees>
    <Employee FirstName="John" LastName="Summers" PhoneNumber="703-548-7841" Department="Finance"></Employee>
    <Employee FirstName="Susan" LastName="Hughey" PhoneNumber="549-461-7962" Department="HR"></Employee>

那么，如果 Department 是一个复杂的对象并且它是 XML 文件中的一个字符串，那么如何才能我更改 LoadData() 函数以将其导入到我的 Employee 对象集合中？

Answer 1

您有几个选择，但一切都取决于您获取 XML 的方式（它是如何保存的）。
（在我看来）最简单的阅读方法是：

XmlSerializer serializer = new XmlSerializer(typeof(YourType));
using (TextReader tr = new StreamReader("newSecret.xml"))
{
 YourType rrr = (YourType)serializer.Deserialize(tr);
}

更多示例：
http://msdn.microsoft.com/en-us/library/he66c7f1。 aspx

另一方面，如果您（出于某种原因）应该使用 LINQ - 请看这里：
LINQ to XML：创建复杂匿名类型
在这里
http://blogs.msdn.com/b/xmlteam/archive/2007/03/24/streaming-with-linq-to-xml-part-2.aspx
希望有帮助！

Answer 2

如果您可以在加载员工之前获取所有部门的列表，则可以将部门放入 Dictionary 中，其中键是部门名称。然后，您可以为每个员工加载正确的部门：

var departmentsDict = departments.ToDictionary(d => d.Name);

XDocument employeesDoc = XDocument.Load("Employees.xml");
List<Employee> data = (from employee in employeesDoc.Descendants("Employee")
   select new Employee
   {
      FirstName= employee.Attribute("FirstName").Value,
      LastName = employee.Attribute("LastName ").Value,
      PhoneNumber = employee.Attribute("PhoneNumber").Value,
      Department = departmentsDict[employee.Attribute("Department").Value]
   }).ToList();
Employees.ItemsSource = data;

代码可能需要修改，具体取决于您在该部门不存在或某人没有指定任何部门时要执行的操作。这段代码在这两种情况下都会抛出异常。

Answer 3

    using System;
using System.Collections.Generic;
using System.IO;
using System.Text;
using System.Xml;
using System.Xml.Serialization;

namespace ConsoleApplication2
{

    public class Employee
    {
        public int Id { get; set; }
        public string FirstName { get; set; }
        public string LastName { get; set; }
        public string PhoneNumber { get; set; }
        public Department Department { get; set; }
    }
    public class Department
    {
        public int Id { get; set; }
        public string Name { get; set; }
        public string Description { get; set; }
        public Employee Manager { get; set; }
    }


    internal class Program
    {
        private static void Main(string[] args)
        {
            String filepath = @"C:\\rrrr.xml";

            #region Create Test Data
            List<Employee> list = new List<Employee>();
            for (int i = 0; i < 5; i++)
            {
                list.Add(new Employee
                             {
                                 Department = new Department
                                                  {
                                                      Description = "bla bla description " + i,
                                                      Id = i,
                                                      Manager = null,
                                                      Name = "bla bla name " + i
                                                  },
                                 FirstName = "First name " + i,
                                 Id = i + i,
                                 LastName = "Last name " + i,
                                 PhoneNumber = Guid.NewGuid().ToString()
                             });
            } 
            #endregion

            #region Save XML
            XmlSerializer serializer = new XmlSerializer(typeof(List<Employee>));
            using (Stream fs = new FileStream(filepath, FileMode.Create))
            {
                using (XmlWriter writer = new XmlTextWriter(fs, Encoding.Unicode))
                {
                    serializer.Serialize(writer, list);
                }
            } 
            #endregion


            //Read from XML

            XmlDocument doc = new XmlDocument();
            doc.Load(filepath);

            List<Employee> newList = new List<Employee>();
            foreach (XmlNode node in doc.GetElementsByTagName("Employee"))
            {
                Employee ee = GetEmploee(node);
                newList.Add(ee);
            }

            //ta da
        }

        public static Employee GetEmploee(XmlNode node)
        {
            return node == null
                       ? new Employee()
                       : new Employee
                             {
                                 Department = GetDepartment(node["Department"]),
                                 FirstName = (node["FirstName"]).InnerText,
                                 LastName = (node["LastName"]).InnerText,
                                 Id = Convert.ToInt32((node["Id"]).InnerText),
                                 PhoneNumber = (node["PhoneNumber"]).InnerText
                             };
        }

        public static Department GetDepartment(XmlNode node)
        {
            return node == null
                       ? new Department()
                       : new Department
                             {
                                 Description = node["Description"].InnerText,
                                 Id = Convert.ToInt32(node["Id"].InnerText),
                                 Manager = GetEmploee(node["Manager"]),
                                 Name = node["Name"].InnerText
                             };
        }
    }
}