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如果我尝试先使用代码,用户类:

public class User
{
    [Key]
    public Guid ID { get; set; }

    [Required(ErrorMessage = "You must pick a user name")]
    [DisplayName("User Name")]
    public string Username { get; set; }

    [Required]
    public string FullName { get; set; }

    [Required(ErrorMessage = "You give a valid email address")]
    [DataType(DataType.EmailAddress)]
    public string Email { get; set; }
    [Required]
    [DataType(DataType.Password)]
    public string Password { get; set; }        
    [Required]
    public DateTime CreateAt { get; set; }

    public string Image { get; set; }

    public bool RemeberMe { get; set; }

    public bool Dummy { get; set; }

    public virtual ICollection<UserFriend> Friends { get; set; }
 } 

UserFriend 类: - 持有用户之间的关系

public class UserFriend
{
    [Key, Column(Order = 0), ForeignKey("User")]
    public Guid UserId { get; set; }
    public User User { get; set; }

    [Key, Column(Order = 1), ForeignKey("Friend")]
    public Guid FriendId { get; set; }
    public User Friend { get; set; }

    public DateTime DateCreated { get; set; }

}

用户存储库:

public class UserRepository : IUserRepository
{
    public User GetUserByName(string userName)
    {
        return context.Users.FirstOrDefault(u => u.Username == userName);
    }
}

如果我尝试使用这种方法:

var userRepo = new UserRepository(new DataContext());
User user = userRepo.GetUserByName(User.Identity.Name); 

问题:如果我做对了,user.Friends.first().Friend 返回 null 我的延迟加载不起作用。

请帮忙...

项目链接 - https://github.com/RanDahan/barker

4

1 回答 1

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User类的构造函数中,您应该创建Friends集合的新实例

Friends = new HashSet<UserFriend>()

另外,我认为UserFirends中的导航属性也应该被声明为虚拟的。

于 2012-06-16T08:23:53.763 回答