Linq to SQL外键 [英] Linq to SQL Foreign Keys

问题描述

用于数据库表的DDL:

DDL for Database Tables:

    Users:
    id - int - identity
    name - varchar - unique

    PCs:
    id - int - idnetity
    name - varchar - unique
    userid - FK to Users

    Apps:
    id - int - identity
    name - varchar
    pcid - FK to PCs

我在Visual Studio 2008中使用Linq To SQL设计器创建了一个DataContext.

I created a DataContext using the Linq To SQL designer in Visual Studio 2008.

我要执行此查询:

select
  users.name,
  pcs.name,
  apps.name
from
  users u
  join pcs p on p.userid = u.id
  join apps a on a.pcid = p.id

在另一个线程中，我被告知我在其中发布了一个答案，指出以下内容不正确，并创建了交叉联接.

I was told in another thread where I posted an answer that the following was incorrect and that it created a cross-join.

var query = from u in db.Users // gets all users
        from p in u.PCs // gets all pcs for user
        from a in p.Apps // gets all apps for pc
        select new
        {
            username = u.Name,
            pcname = p.Name,
            appname = a.Name
        };

当我执行此查询时，我得到正确的结果.在每个表中有两个记录的交叉联接应该返回8个记录，但是我的查询正确地返回了两个记录.

When I execute this query I get the correct results. A cross-join with two records in each table should return 8 records but my query correctly returns the two records.

我很幸运吗，还是那个人告诉我我错了吗?

Am I lucky, or is the person telling me that I'm wrong confused?

推荐答案

这将起作用.这是正常现象，因为第二个"from"查询个人计算机的用户组，第三个"from"询问个人计算机的应用程序组.而且我想这不会像T-Sql那样产生交叉联接，因为条件已经在外键和关联中定义了.

That will work. It's normal because the second "from" inquires in the PC's set of the users and the third "from", the Apps set of the pcs. And I guess that is not going generate a cross join as T-Sql, because the conditions are already defined in the foreign keys and associations.

但是我想这应该是您希望通过内部联接获得的语法；

But I guess this should be the syntactically you want by inner join;

var query = from u in db.Users
            join p in db.PCs on p.UserId == u.Id
            join a in db.Apps on a.PCId == p.Id
            select new
            {
                username = u.Name,
                pcname = p.Name,
                appname = a.Name
            };

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