I have no idea how I can implement this quite simple task using NHibenate 1.2.0.

I have a business entity Order with a collection of child objects OrderItem.

Usually I use Orders without OrderItems, so I set lazy=true for them. But sometimes I need to load a list of Orders with related OrderItems at once with maximum performance.
How can do it without enumerating of all child objects (to enable lazy loading)? Maybe I can force NHibernate to load all child order items during loading of orders (simulate lazy=false behavior)?

Hi Almas,

you may do this by setting your lazy option to true, or not specifiing it (since in NHibernate 1.2.0 the lazy option defaults to true).

In your code you load your entity Order like you are used to to.
If you get an order for witch you'll need all the childs you can load them by performing a:
NHibernateUtil.Initialize(Order[x].OrderItems);

regards
Stephan

Thank you, Stephan.
But, should I call NHibernateUtil.Initialize for every order in the list?
Does it produce an additional select of items for every order?

Do NOT call:
NHibernateUtil.Initialize(Order[x].OrderItems);

This will create a SELECT N+1 issue, use eager loading, like this:

session
.CreateQuery("select o from Order o left join fetch o.OrderLines where o.Id = :id")
.SetInt32("id", 5)
.UniqueResult<Order>();

This will load the order with all its OrderLines already populated

If you don't want to use HQL, you can still use Nhibernate.Initialize(...).

The select N+1 problem can be avoided by using the "batch-size" tag, or using the fetch="subselect" mode, which is new in the latest beta.

This query produces a set of duplicated orders which number is proportional to the number of its related order lines. Is it strange behaviour by design?

It is behavior by design. The order objects are actually just duplicate references, not duplicate objects. Put all of the orders into a Set to remove the duplicates.

Cheers,

Symon.

Thank you, guys.

Are there any references to a documentation or articles that explain that behavior?