序列化后只保存最新的客户

如何解决序列化后只保存最新的客户

我需要构建一个控制台应用程序，允许用户输入（通过终端）新用户的详细信息。然后需要通过序列化将其写入 XML（必需）。我构建了我的客户类并有一个方法来构建新用户 - 但它总是会忘记最后一个条目并且只将一个用户实例写入我的列表。

这是我为添加用户而构建的方法：

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Xml.Serialization;


namespace Hausarbeit_Autovermietung_Gierow
{
    class Addcustomer
    {
        public void AddCustomer()
        {
            
            var customer = new Customer();
            List<Customer> customers = new List<Customer>() { customer };
            //DeserializeFromXML(customer);
            DeserializeFromXML(customers);
            var listCount = customers.Count;

            int maxID = FindMaxValue(customers,x => x.ID);

            Console.WriteLine("Vorname eingeben");
            customer.Firstname = Console.ReadLine();
            Console.WriteLine("Nachname eingeben");
            customer.Lastname = Console.ReadLine();
           // int ID;
            customer.ID = maxID + 1;
            
            //List<Customer> customers = new List<Customer>() { customer };

            SerializeToXML(customer);
            SerializeToXML(customers);
            
            foreach (var cust in customers)
            {
                Console.WriteLine($"Vorname: {cust.ID} {cust.Firstname} {cust.Lastname}");


                Console.ReadKey();

            }
            }

        static public void SerializeToXML(Customer customer)
        {
            XmlSerializer serializer = new XmlSerializer(typeof(Customer));
            using (TextWriter textWriter = new StreamWriter(@"customer.xml"))
            {
                serializer.Serialize(textWriter,customer);
            }
        }

        public static void SerializeToXML(List<Customer> customers)
        {
            //var customer = new Customer("Vorname","Nachname");
            XmlSerializer serializer = new XmlSerializer(typeof(List<Customer>));
            using (System.IO.TextWriter textWriter = new System.IO.StreamWriter(@"List.xml"))
            {
                serializer.Serialize(textWriter,customers);
            }

        }

       static List<Customer> DeserializeFromXML(List<Customer> customers)
        {
            XmlSerializer deserializer = new XmlSerializer(typeof(List<Customer>));
            List<Customer> customerslist;
            using (TextReader textReader = new StreamReader(@"List.xml"))
            {
                customerslist = (List<Customer>)deserializer.Deserialize(textReader);
                return customerslist;
            }
        }

        public int FindMaxValue<T>(List<T> list,Converter<T,int> projection)
        {
            if (list.Count == 0)
            {
                throw new InvalidOperationException("Empty list");
            }
            int maxValue = int.MinValue;
            foreach (T item in list)
            {
                int value = projection(item);
                if (value > maxValue)
                {
                    maxValue = value;
                }
            }
            return maxValue;
        }
    }
}

解决方法

在这里创建一个仅包含新客户的列表：

List<Customer> customers = new List<Customer>() { customer };

然后通过反序列化 xml 使用新实例覆盖它：

DeserializeFromXML(customers);

但您永远不会添加新客户（添加此行）：

customers.Add(customer);

因此，当您进行序列化时，新客户将成为列表的一部分：

SerializeToXML(customers);