ADO.NET Entity Framework 4.1 C# DAL 创建/添加对象并按键访问

Question

我创建了一个 DAL 类“FriendManager”，封装了对我的实体的访问。

public class FriendManager
{
  private readonly FriendsEntities _db;
  public FriendManager()
  {
    _db = new FriendsEntities();
  }

  public void SaveChanges()
  {
    _db.AcceptAllChanges();
  }

  public Guid CreateFriend(string forename, string surname)
  {
    var id = Guid.NewGuid();
    _db.Friends.AddObject(new Friend { Identifier = id, Name = string.Concat(forename, , " ", surname) });
    _db.SaveChanges(SaveOptions.DetectChangesBeforeSave);
    return id;
  }

  public void AddFriendsAddress(Guid friendId, string address)
  {
    var friend = GetFriend(friendId);
    friend.Address = address;
    _db.SaveChanges(SaveOptions.DetectChangesBeforeSave);
  }

  public Friend GetFriend(Guid id)
  {
    return (from friends in _db.Friends
            where friends.Identifier == id
            select friends).SingleOrDefault();
  }
}

现在我想在控制台应用程序中创建一个朋友，然后通过在调用 SaveChanges() 之前立即访问它来将创建的条目写入命令行！

using (var manager = new FriendManager())
{
  bool success = false;
  try
  {
    using (var scope = new TransactionScope())
    {
      var id = manager.CreateFriend("John", "Fischer");
      manager.AddFriendsAddress(id, "Blue Hole 12, 1234 Atlantic, World");
      var john = manager.GetFriend(id);
      Console.WriteLine("My new friend {0} lives in {1}.", john.Name, john.Address);
      scope.Complete();
    }
    success = true;
  }
  catch(Exception ex)
  {
    Console.WriteLine("The following went wrong: {0}", ex.Message);        
  }
  if (success) manager.SaveChanges();
}

Console.WriteLine("Press <ENTER> to exit...");
Console.ReadLine();

现在我收到 PK 违规错误，因为 EF 试图将 John 两次添加到我的 Friends 实体中！如果我不调用“_db.SaveChanges(SaveOptions.DetectChangesBeforeSave);”在“CreateFriend”方法中，则“AddFriendsAddress”失败，因为在 Friends 实体中找不到 John（“GetFriend”的返回值始终为 NULL）！

Answer 1

尝试将您的控制台应用程序修改为：

using (var manager = new FriendManager())
{
var id = manager.CreateFriend("John", "Fischer");
manager.SaveChanges();
var john = manager.GetFriend(id);
Console.WriteLine("My new friend {0}.", john.Name);
}

您应该先将数据保存到数据库，然后按键调用它。