asp.net-mvc - Entity Framework - 无法创建类型为 x 的常量值。在此上下文中仅支持原始类型。

Question

我有以下

工程师模型:

public class engineers
{
    public Guid? Guid { get; set; }
    public string Name { get; set; }
}

我用正确的详细信息填写工程师列表:

List<engineers> listeng = new List<engineers>();
listeng.Add(new engineers { Name = profile.FirstName + " " + profile.LastName, Guid = GuidEngineer });

到现在为止还挺好。

我的问题是如何将工程师姓名拉到下面的 eng 条目中:

 var tickets = from o in new HelpdeskEntities().Tickets.Where(t => t.TicketState.State == "Open")
                    select new AjaxTickets
                    {
                        TicketID = o.TicketID,
                        TicketSubject = o.TicketSubject,
                        ClientCompanyName = o.ClientCompany.ClientCompanyName,
                        DateOpened = o.DateOpened,
                        **eng** = list.Where(x => x.Guid == o.EngineerID).Select(x => new engineers {Guid = x.Guid, Name=x.Name }).FirstOrDefault().Name

                    }; 

我也试过

var tickets = from o in new HelpdeskEntities().Tickets.Where(t => t.TicketState.State == "Open")
                    select new AjaxTickets
                    {
                        TicketID = o.TicketID,
                        TicketSubject = o.TicketSubject,
                        ClientCompanyName = o.ClientCompany.ClientCompanyName,
                        DateOpened = o.DateOpened,
                        **eng** = list.Where(x => x.Guid == o.EngineerID).Select(x => x.Name }).FirstOrDefault()
                    }; 

我得到的错误是:

Unable to create a constant value of type 'Helpdesk2.ViewModel.engineers'. Only primitive types ('such as Int32, String, and Guid') are supported in this context."}

我有点理解但无法弄清楚只是为了选择工程师名称。

提前致谢

Answer 1

您应该能够将第一个简化为:

list.FirstOrDefault(x => x.Guid == o.EngineerID).Name

话虽如此， Entity Framework 可能不会让您在运行数据库调用时运行它。如果你可以制作一个从票证到工程师的外键，那么你可以像选择客户公司名称一样选择它。如果没有，那么您需要分两步完成:首先，在不填充工程师姓名属性的情况下运行选择，然后用以下内容填充它们:

tickets.ForEach(ticket => ticket.EngineerName = engineers.First(eng => eng.Guid == ticket.EngineerID).Name)

显然，您需要添加 EngineerID 属性并在第一步中选择它。