Question

我想通过实体框架绑定gridview但它会抛出错误像 -

匿名类型不能具有同名Entity Framwrok
的多个属性

这是我的方法。

public void UserList(GridView grdUserList)
{
    using (TreDbEntities context = new TreDbEntities())
    {

        var query =( from m in context.aspnet_Membership
                    from u in context.aspnet_Users
                    join usr in context.Users
                    on new { m.UserId, u.UserId } 
                    equals new { usr.MembershipUserID, usr.UserId }
                    into UserDetails
                    from usr in UserDetails
                    select new { 
                       CreationDate = m.CreateDate,
                       email = m.Email,
                       UserName = u.LoweredUserName,
                       Name = usr.FirstName + usr.LastNameLastName,
                       Active=usr.IsActive
                    }).ToList();
    }
}

这里显示错误。 usr.UserId。

Answer 1

直接问题是匿名类型new { m.UserId, u.UserId }：两次同名。您可以通过提供显式属性名称来解决此问题，例如：new { u1 = m.UserId, u2 = u.UserId }。

但接下来的问题是，定义连接的两个匿名类型都不会具有相同的属性名称，因此最终修复是这样的：

public void UserList(GridView grdUserList)
{
    using (TreDbEntities context = new TreDbEntities())
    {
        var query =( from m in context.aspnet_Membership
                    from u in context.aspnet_Users
                    join usr in context.Users
                    on new { u1 = m.UserId, u2 = u.UserId } 
                    equals new { u1 = usr.MembershipUserID, u2 = usr.UserId }
                    into UserDetails
                    from usr in UserDetails
                    select new { CreationDate = m.CreateDate,
                                 email = m.Email,
                                 UserName = u.LoweredUserName,
                                 Name = usr.FirstName + " " + usr.LastName,
                                 Active = usr.IsActive
                               }
                   ).ToList();
    }
}

Answer 2

@Gert回答是正确的。只想显示更简单的解决方案 - 仅为第一个UserId属性命名：

on new { MembershipUserID = m.UserId, u.UserId }

匿名类型不能具有多个具有相同名称的属性

2 个答案: