我有点困惑。直到今天,我认为必须在DbContext
类中指定每个表(由EF使用)。但看起来我只需要一个!真?
让我解释一下,这是我的DbContext:
public class MyDbContext : DbContext
{
public MyDbContext()
: base("name=MyDbContext")
{
}
protected override void OnModelCreating(DbModelBuilder modelBuilder)
{
Database.SetInitializer<MyDbContext>(null);
base.OnModelCreating(modelBuilder);
}
public DbSet<Table1> Table1 { get; set; }
public DbSet<Table2> Table2 { get; set; }
public DbSet<Table3> Table3 { get; set; }
public DbSet<Table4> Table4 { get; set; }
public DbSet<Table5> Table5 { get; set; }
}
这是两个示例表,连接1:很多
[Table("Table1")]
public class Table1
{
[Key]
[Column("Table1Id", TypeName = "uniqueidentifier")]
public int Table1Id { get; set; }
[Column("Table2Id", TypeName = "int")]
public int Table2Id { get; set; }
[ForeignKey("Table2Id")]
public Table2 Table2 { get; set; }
}
[Table("Table2")]
public class Table2
{
public Table2()
{
this.Table1s = new HashSet<Table1>();
}
[Key]
[Column("Table2Id", TypeName = "int")]
[DatabaseGenerated(DatabaseGeneratedOption.Identity)]
public int Table2Id { get; set; }
public ICollection<Table1> Table1s { get; set; }
}
简单。现在,我想查询所有Table2s与相应的Table1s。我做:
var tables2 = fni.Set<Table2>()
.Include(i => i.Table1s)
.Where(t => t.Table2Id == 123456).ToList();
这一切都有效,但当我偶然发现它甚至可以使用这个DbContext时,我很震惊:
public class MyDbContext : DbContext
{
public MyDbContext()
: base("name=MyDbContext")
{
}
protected override void OnModelCreating(DbModelBuilder modelBuilder)
{
Database.SetInitializer<MyDbContext>(null);
base.OnModelCreating(modelBuilder);
}
public DbSet<Table1> Table1 { get; set; }
}
或这个..
public class MyDbContext : DbContext
{
public MyDbContext()
: base("name=MyDbContext")
{
}
protected override void OnModelCreating(DbModelBuilder modelBuilder)
{
Database.SetInitializer<MyDbContext>(null);
base.OnModelCreating(modelBuilder);
}
public DbSet<Table2> Table2 { get; set; }
}
你能解释一下,为什么它有效?怎么样?
谢谢!
编辑。它不包括在内。我能做到:
var tables2 = fni.Set<Table2>()
.Where(t => t.Table2Id == 123456).ToList();
只有这个:在DbContext中的public DbSet<Table1> Table1 { get; set; }
。它甚至不是Table2!它们通过FK连接(定义没有改变)。这意味着,您必须只有一个表“链”中的一个表。那是对的吗?