Question

我目前正在与XMLSerializer一起玩，以了解其工作原理。我能够序列化，保存和反序列化单个对象而不会出现问题。但是，当我尝试反序列化多个对象时遇到了问题。我收到此错误：Unhandled exception. System.InvalidOperationException: There is an error in XML document (10, 10). ---> System.Xml.XmlException: Unexpected XML declaration. The XML declaration must be the first node in the document, and no whitespace characters are allowed to appear before it.

我已经尝试过这种方法https://stackoverflow.com/a/16416636/8964654 在这里（我可能做错了）


 public static ICollection<T> DeserializeList<T>()
    {


      string filePath = @"TextFiles/Users.txt";
      XmlSerializer serializerTool = new XmlSerializer(typeof(User));
             List<T> list = new List<T>();


      using (FileStream fs = new FileStream (filePath, FileMode.Open)){

       while(fs.Position!=fs.Length)
       {
         //deserialize each object in the file
         var deserialized = (T)serializerTool.Deserialize(fs); 
         //add individual object to a list
         list.Add(deserialized);
        }
      }

    //return the list of objects
    return list;
}

它不起作用

这是我的原始代码。我特意调用了SaveUser方法两次，以模拟该方法在不同时间被调用两次

 [Serializable]
  public class User: ISerializable{

    public static void SaveUser(User user){
      string filePath = @"TextFiles/Users.txt";
      XmlSerializer serializerTool = new XmlSerializer(typeof(User));

      using(FileStream fs = new FileStream(filePath, FileMode.Append)){
        serializerTool.Serialize(fs, user);
        }
    }

    public static void PrintUser(){
      string filePath = @"TextFiles/Users.txt";
      XmlSerializer serializerTool = new XmlSerializer(typeof(User));

      using (FileStream fs = new FileStream (filePath, FileMode.Open)){
        User u1 = (User)serializerTool.Deserialize(fs);
        Console.WriteLine($"{u1.FirstName} {u1.LastName}, {u1.DOB.ToShortDateString()}");
        }
    }
}


class Program
    {
        static void Main(string[] args)
        {

    User user1 = new User(){
      FirstName = "Kim",
      LastName = "Styles",
      Address = "500 Penn street, Dallas, 46589",
      Username = "KimStyles@yahoo.com",
      Password ="Kim2019",
      DOB = (new DateTime(1990,10,01)),
      Id = 2
    };


     User user2 = new User(){
      FirstName = "Carlos",
      LastName = "Santana",
      Address = "500 Amigos street,San Jose, California, 46589",
      Username = "Carlos.Santana@yahoo.com",
      Password ="CarLosSan2019",
      DOB = (new DateTime(1990,10,01)),
      Id = 2
    };

   User.SaveUser(user1);
   User.SaveUser(user2);
   User.PrintUser();

        }
    }

下面是它如何保存XML数据


<?xml version="1.0"?>
<User xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <FirstName>Kim</FirstName>
  <LastName>Styles</LastName>
  <DOBProxy>Monday, 01 October 1990</DOBProxy>
  <Username>KimStyles@yahoo.com</Username>
  <Password>Kim2019</Password>
  <Address>500 Penn street, Dallas, 46589</Address>
  <Id>1</Id>
</User>
<?xml version="1.0"?>
<User xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <FirstName>Carlos</FirstName>
  <LastName>Santana</LastName>
  <DOBProxy>Monday, 01 October 1990</DOBProxy>
  <Username>Carlos.Santana@yahoo.com</Username>
  <Password>CarLosSan2019</Password>
  <Address>500 Amigos street,San Jose, California, 46589</Address>
  <Id>2</Id>
</User>

我希望能够检索每个用户的所有数据并打印详细信息。我怎样才能做到这一点？有更好的方法吗？

Answer 1

您的xml具有多个根元素，有效的xml不允许这样做。如果将其更改为格式，则应该可以。

<?xml version="1.0"?>
<Users>
   <user></user>
   <user></user>
</Users>

Answer 2

我将按照以下方式解决此问题：

创建 User 类

可序列化类包含用户详细信息。

response.getWriter().append("<table><body>");
        for (int i = 1; i < width +1 ; i++) {
            response.getWriter().append("<tr>");
            for (int j = 1; j < height +1; j++) {
                response.getWriter().append("<td>").append(String.format("%4d",i*j)).append("</td>");
            }
            response.getWriter().append("</tr>");
        }
        response.getWriter().append("</body></table>");

创建 Users 类

另一个 Serializable 类包含[Serializable] public class User { public int ID { get; set; } public string FirstName { get; set; } public string LastName { get; set; } public DateTime DOB { get; set; } public override string ToString() { return $"{ID}, {FirstName}, {LastName}, {DOB.ToShortDateString()}"; } }对象的列表，并处理序列化和反序列化例程：

User

这样，您可以确保XML文件的格式正确，管理用户列表（添加，删除，编辑）。

保存（序列化）示例

[Serializable]
public class Users  
{
    public List<User> ThisUsers = new List<User>();

    public void Save(string filePath)
    {
        XmlSerializer xs = new XmlSerializer(typeof(Users));

        using (StreamWriter sr = new StreamWriter(filePath))
        {
            xs.Serialize(sr, this);
        }
    }

    public static Users Load(string filePath)
    {
        Users users;
        XmlSerializer xs = new XmlSerializer(typeof(Users));
        using (StreamReader sr = new StreamReader(filePath))
        {
            users = (Users)xs.Deserialize(sr);
        }
        return users;
    }
}

加载（反序列化）示例：

string filePath = @"TextFiles/Users.txt";
Users users = new Users();
for (int i = 1; i < 5; i++)
{
    User u = new User
    {
        ID = i,
        FirstName = $"User {i}",
        LastName = $"Last Name {i}",
        DOB = DateTime.Now.AddYears(-30 + i)                    
    };
    users.ThisUsers.Add(u);
}
users.Save(filePath);

这是生成的XML文件的外观

string filePath = @"TextFiles/Users.txt";
Users users = Users.Load(filePath);
users.ThisUsers.ForEach(a => Console.WriteLine(a.ToString()));

//Or get a specific user by id:
Console.WriteLine(users.ThisUsers.Where(b => b.ID == 3).FirstOrDefault()?.ToString());

祝你好运。

无法序列化和反序列化多个对象

2 个答案: