asp.net 从 FindAll 或 Where 修改列表而不修改搜索的原始列表

Question

我正在尝试从 findall 或 where 创建一个新列表并修改该新列表而不更改原始列表。从 FindAll 创建 list2 并对 list2 进行更改时，它不应影响 list1。怎么会？

编辑...使用了似乎有效的两个答案的组合。有没有预见到的问题？？？工作代码编辑 编辑...它不起作用。作为对象属性的类没有复制其数据......有什么想法吗？？？我不明白为什么只是复制或克隆对于列表来说并不简单。

class Program
{
    static void Main(string[] args)
    {
        Console.WriteLine("Started");
        List<SampleData> list1 = SampleData.MakeList(3);
        for (int i = 0; i < list1.Count; i++)
        {
            Console.WriteLine("list1.Number => " + list1[i].Number);
            Console.WriteLine("list1.Person => " + list1[i].Person);
        }
        List<SampleData> list2 = new List<SampleData>(list1.FindAll(m => m.Number == 1).Select(x => new SampleData().ShallowCopy())).ToList();
        for (int i = 0; i < list2.Count; i++)
        {
            list2[i].Person = "Person " + list2[i].Number;
            Console.WriteLine("list2.Number => " + list2[i].Number);
            Console.WriteLine("list2.Person => " + list2[i].Person);
        }

        SampleData sd = list2.Find(s => s.Number == 1);
        Console.WriteLine("Apartment Number" + sd.apartment.BuildingLetter); // <<<<------------------- THROWING NULL EXCEPTION FOR APARTMENT

        Console.WriteLine("-------AFTER MODIFYING NEW LIST----LIST1.Person SHOULD BE NULL---");
        for (int i = 0; i < list1.Count; i++)
        {
            Console.WriteLine("--list1.Number => " + list1[i].Number);
            Console.WriteLine("--list1.Person => " + list1[i].Person);
        }
        for (int i = 0; i < list2.Count; i++)
        {
            Console.WriteLine("--list2.Number => " + list2[i].Number);
            Console.WriteLine("--list2.Person => " + list2[i].Person);
        }
        Console.ReadLine();
        System.Environment.Exit(0);
    }

}
class SampleData
{      
    public int Number { get; set; }
    public String Person { get; set; }
    public Apartment apartment { get; set; }
    public static List<SampleData> MakeList(int count)
    {
        List<SampleData> l = new List<SampleData>();
        for (int i = 0; i < count; i++)
        {
            SampleData d = new SampleData();
            d.apartment = new Apartment { RoomNumber = i, BuildingLetter = "letter-" + i };
            d.Number = i;
            l.Add(d);
        }
        return l;
    }
    public SampleData ShallowCopy()
    {
        return (SampleData)this.MemberwiseClone();
    }

}
class Apartment
{
    public int RoomNumber { get; set; }
    public String BuildingLetter { get; set; }
}

Answer 1

ShallowCopy 是你的朋友。

像这样在 SampleData 类中定义一个 ShallowCopy() 方法。

class SampleData
{
    public int Number { get; set; }
    public String Person { get; set; }
    public static List<SampleData> MakeList(int count)
    {
        List<SampleData> l = new List<SampleData>();
        for (int i = 0; i < count; i++)
        {
            SampleData d = new SampleData();
            d.Number = i;
            l.Add(d);
        }
        return l;
    }

    public SampleData ShallowCopy()
    {
        return (SampleData)this.MemberwiseClone();
    }

}

然后按如下方式填充list2...

List<SampleData> list2 = new List<SampleData>();
foreach (SampleData sd in list1.Where(m => m.Number == 1))
{
    list2.Add(sd.ShallowCopy());
}

Answer 2

您可以使用 。 AsNoTracking() 方法在从 DbContext 或 ObjectContext

中选择时 Entity Query

List<SampleData> list2 = new List<SampleData>(list1.Where(m => m.Number == 1).AsNoTracking().ToList());

---

或使用类似的复制承包商

class SampleData
{
    public int Number { get; set; }
    public String Person { get; set; }

    public SampleData(SampleData obj)
    {
        This.Number=obj.Number;
        This.Person=obj.Person;
        ..
        ..
    }
    public static List<SampleData> MakeList(int count)
    {
        List<SampleData> l = new List<SampleData>();
        for (int i = 0; i < count; i++)
        {
            SampleData d = new SampleData();
            d.Number = i;
            l.Add(d);
        }
        return l;
    }
}

然后

 List<SampleData> list2 = new List<SampleData>(list1.Where(m => m.Number == 1).Select(x=>new SampleData(x)).ToList());