通过两个非主键列连接的流畅 Nhibernate 映射

Question

我想知道是否有一种方法可以在不创建视图的情况下连接两个具有两个非主键列的表？我有一个名为“Make”的表，其中包含“Name”和“Year”列，我想将其与另一个名为“Style”的表连接，其中包含“MakeName”和“MakeYear”列。一种“品牌”可以有多种“风格”。 以下是我迄今为止创建的实体：

public class Make
{
    public virtual int Id { get; set; }
    public virtual string Name { get; set; }
    public virtual string Year { get; set; }
    public virtual IList<Style> Styles { get; set; } 
}

public class Style
{
    public virtual int Id { get; set; }
    public virtual string MakeName { get; set; }
    public virtual string MakeYear { get; set; }
    public virtual string Class { get; set; }
    public virtual Make Make { get; set; }
}

另外，这些是我迄今为止拥有的类映射：

public class MakeMap : ClassMap<Make>
{
    public MakeMap()
    {
        Id(x => x.Id);
        Map(x => x.Name);
        Map(x => x.Year);
        // Bad mapping...
        //HasManyToMany(x => x.Styles).Table("Make").AsBag()
          .ParentKeyColumn("MakeName")
          .ChildKeyColumn("MakeYear").Cascade.All();
        Table("Make");
    }
}

public class StyleMap : ClassMap<Style>
{
    public StyleMap()
    {
        Id(x => x.Id);
        Map(x => x.Class);
        Map(x => x.MakeName);
        Map(x => x.MakeYear);
        // Ends up overwriting the "MakeName" column
        References(x => x.Make).Column("MakeName").PropertyRef("Name").
              Column("MakeYear").PropertyRef("Year");
        Table("Style");
    }
}

谢谢！

Answer 1

3 个选项，我更喜欢第一个

1. 选项：

   更改数据库架构以将make id而不是名称和年份包含到样式表中

2. Option

References(x => x.Make)
    .Formula("(Select s.Id FROM Style s WHERE s.Name = MakeName AND s.Year = MakeYear)");

3. 选项

public class Make
{
    public virtual int Id { get; set; }

    public virtual MakeId BusinessId { get; private set; }
}

public class MakeId
{
    public virtual string Name { get; set; }
    public virtual string Year { get; set; }
}

public class MakeMap : ClassMap<Make>
{
    public MakeMap()
    {
        Id(x => x.Id);
        Component("BusinessId", c => 
            c.Map(x => x.Name);
            c.Map(x => x.Year);
        });

        HasMany(x => x.Styles)
          .PropertyRef("BusinessId")
          .KeyColumns.Add("MakeName", "MakeYear")
          .Cascade.All();
        Table("Make");
    }
}

public class StyleMap : ClassMap<Style>
{
    public StyleMap()
    {
        Table("Style");

        Id(x => x.Id);
        Map(x => x.Class);

        References(x => x.Make)
            .PropertyRef("BusinessId")
            .Columns.Add("MakeName", "MakeYear");
    }
}