CODEWITHSUNDEEP
c#xmlserializationdeserializationfilestream
user8964654
user8964654

Reputation: 97

Trouble serializing and deserializing multiple objects

I currently playing with the XMLSerializerto understand how it works. I am able to serialize, save and de-serialize a single object without problem. However I run into problems when I try to de-serialize multiple objects. I get this error : 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.

I've tried this approach https://stackoverflow.com/a/16416636/8964654 here (and I could be doing it wrong)


 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;
}

it didn't work

This is my original code. I intentionally call the SaveUser method twice to simulate the method being called twice at different times

 [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 = "[email protected]",
      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 = "[email protected]",
      Password ="CarLosSan2019",
      DOB = (new DateTime(1990,10,01)),
      Id = 2
    };

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

        }
    }

below is how it saved XML data


<?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>[email protected]</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>[email protected]</Username>
  <Password>CarLosSan2019</Password>
  <Address>500 Amigos street,San Jose, California, 46589</Address>
  <Id>2</Id>
</User>

I want to be able to retrieve all the data and print details of each individual user. How can I do this? Is there a better approach?

Upvotes: 0

Views: 448

Answers (2)

user10216583
user10216583

Reputation:

I'd solve this problem as follow:

Create the User class

A Serializable class contains a user details.

[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()}";
    }
}

Create the Users class

Another Serializable class contains a list of User objects and handles both serialize and Deserialize routines:

[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;
    }
}

This way, you guarantee the XML file is formatted correctly, manage the users list (add, remove, edit).

Save (serialize) example

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);

Load (Deserialize) example:

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());

and here is how the generated XML file looks like

<?xml version="1.0" encoding="utf-8"?>
<Users xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
  <ThisUsers>
    <User>
      <ID>1</ID>
      <FirstName>User 1</FirstName>
      <LastName>Last Name 1</LastName>
      <DOB>1990-11-04T08:16:09.1099698+03:00</DOB>
    </User>
    <User>
      <ID>2</ID>
      <FirstName>User 2</FirstName>
      <LastName>Last Name 2</LastName>
      <DOB>1991-11-04T08:16:09.1109688+03:00</DOB>
    </User>
    <User>
      <ID>3</ID>
      <FirstName>User 3</FirstName>
      <LastName>Last Name 3</LastName>
      <DOB>1992-11-04T08:16:09.1109688+03:00</DOB>
    </User>
    <User>
      <ID>4</ID>
      <FirstName>User 4</FirstName>
      <LastName>Last Name 4</LastName>
      <DOB>1993-11-04T08:16:09.1109688+03:00</DOB>
    </User>
  </ThisUsers>
</Users>

Good luck.

Upvotes: 1

peeyush singh
peeyush singh

Reputation: 1407

Your xml has multiple root elements, this is not allowed for valid xml. If you change it to the format, this should work.

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

Upvotes: 1

Related Questions