If you have the following relationship:

TableA --(1:n)--> TableB --(1:n)--> TableC

I now want to display all Entities from TableC in a Grid. How do I do that?

I can get a list of TableC for any Entity in TableB: EntityA.EntityCollectionB[x].EntityCollectionC

In SQL it would be simple select whatever from TableC join TableB on TableB.TableBKey=Tablec.TableBKey join TableA on TableA.TableAKey=TableB.TableAKey

And I could of course define a view and scope it, or a stored procedure, but thought I would be able to do it with LLBLGen objects?

EmployeeCollection employees = new EmployeeCollection();
IPredicateExpression filter = new PredicateExpression();
filter.Add(CustomerFields.CustomerId == 1);
IRelationCollection relations = new RelationCollection();
relations.Add(CustomerEntity.Relations.DepartmentEntityUsingCustomerId);
relations.Add(DepartmentEntity.Relations.EmployeeEntityUsingDepartmentId);
employees.GetMulti(filter, relations);

This code should do it. TableA is customer, TableB is department and TableC is Employee

employees is a collection that you can then bind to your grid.

I can try that, perhaps I was thinking about it all wrong.

I already have a TableA entity loaded. It has relations defined in the GUI (and hence code) to TableB. I was hoping I could get there using the relations I had setup in the GUI and was just missing something???

When I look at the relations in the GUI, I have relations Customer <--> Department <--> Employee (which all come from the database of course).

Can the GUI not define an 'Employees' collection off of 'CustomerEntity'? (which is what you setup in code I guess?) I see some of the automatic relationships sometimes have 'via' in them, but do not know how to get that setup.

If you want to do it from the LLBLGen Pro designer, then you may create a Typed List.

So there is no way to have myEntity.Somecollection, where Somecollection is more than one relation away?

If I used a TypedEntity, I would then want to relate that to MyEntity so that I can do the above.

Is that code above the 'way' to do this - I would think it would be a designer option or something?

The designer can detect the following relations:

1- [A] m:1 [B] 2- [A] 1:m [B] 2- [A] m:n [B] (which has the via word in the name) = [A] 1:n [C] && [C] m:1 [B]

There isn't a way to do it in the designer using two 1:m relationships. You could put the code I have in the usercode region to return the collection that you are refering too. If you want to do this then I would implement it as another collection by hand by using other related collections as an example in your TableA entity class.

No problem - I just added a property to my entity to do the relation as you have shown me and works like a champ.

Thanks.