[英]Trouble serializing and deserializing multiple objects
我目前正在使用XMLSerializer
来了解它是如何工作的。 我能够毫无问题地序列化、保存和反序列化单个 object。 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.
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.
我在这里尝试过这种方法https://stackoverflow.com/a/16416636/8964654 (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;
}
它没有用
这是我的原始代码。 我故意调用SaveUser
方法两次来模拟在不同时间被调用两次的方法
[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();
}
}
以下是它如何保存 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?
您的 xml 有多个根元素,这对于有效的 xml 是不允许的。 如果将其更改为格式,则应该可以。
<?xml version="1.0"?>
<Users>
<user></user>
<user></user>
</Users>
我将按如下方式解决此问题:
创建用户class
可序列化的 class 包含用户详细信息。
[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()}";
}
}
创建用户class
另一个Serializable class 包含User
对象列表并处理序列化和反序列化例程:
[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;
}
}
这样,您可以保证 XML 文件格式正确,管理用户列表(添加、删除、编辑)。
保存(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);
加载(反序列化)示例:
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());
这是生成的 XML 文件的样子
<?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>
祝你好运。
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