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是否可以使用单表继承的通用支持,并且仍然能够 FindAll 的基类?

作为一个额外的问题,我也可以使用 ActiveRecordLinqBase<> 吗?我确实喜欢这些查询。

更多细节:假设我定义了以下类:

public interface ICompany
{
    int ID { get; set; }
    string Name { get; set; }
}

[ActiveRecord("companies", 
  DiscriminatorColumn="type", 
  DiscriminatorType="String", 
  DiscriminatorValue="NA")]
public abstract class Company<T> : ActiveRecordBase<T>, ICompany
{
    [PrimaryKey]
    private int Id { get; set; }

    [Property]
    public String Name { get; set; }
}

[ActiveRecord(DiscriminatorValue="firm")]
public class Firm : Company<Firm>
{
    [Property]
    public string Description { get; set; }
}

[ActiveRecord(DiscriminatorValue="client")]
public class Client : Company<Client>
{
    [Property]
    public int ChargeRate { get; set; } 
}

这适用于大多数情况。我可以做这样的事情:

var x = Client.FindAll();

但有时我想要所有的公司。如果我不使用泛型,我可以这样做:

var x = (Company[]) FindAll(Company);
Client a = (Client)x[0];
Firm b = (Firm)x[1];

有没有办法编写一个返回 ICompany 数组的 FindAll,然后可以将其类型转换为它们各自的类型?
就像是:

var x = (ICompany[]) FindAll(Company<ICompany>);
Client a = (Client)x[0];

或者,也许我打算实现通用支持都错了?

4

1 回答 1

0

这个怎么样:

[ActiveRecord("companies", 
  DiscriminatorColumn="type", 
  DiscriminatorType="String", 
  DiscriminatorValue="NA")]
public abstract class Company : ActiveRecordBase<Company>, ICompany {
    [PrimaryKey]
    private virtual int Id { get; set; }

    [Property]
    public virtual String Name { get; set; }
}

[ActiveRecord(DiscriminatorValue="firm")]
public class Firm : Company {
    [Property]
    public virtual string Description { get; set; }
}

[ActiveRecord(DiscriminatorValue="client")]
public class Client : Company {
    [Property]
    public virtual int ChargeRate { get; set; } 
}

var allClients = ActiveRecordMediator<Client>.FindAll();
var allCompanies = ActiveRecordMediator<Company>.FindAll(); // Gets all Companies (Firms and Clients). Same as Company.FindAll();

请注意,您不能只是将您的公司贬低为客户或公司,您需要使用适当的多态性或访问者。请参阅this以获得解释。

于 2010-05-05T19:57:15.233 回答