I was unable to make this do what I want. The code below chops my
default collection, which is not what I want to do. Here's my data
relationship. I have five types of objects: Family, Person, Marriage,
Birth, Death. A Family has a Marriage member and Person contains the
other three (Family, Birth, Death). I want to limit my tabular view to
objects of type Person and still have my facets work for that view and
all the other views. (The other views all show all object types.)

> Brannon King wrote:
> > So the problem with this plan is that when you create a collection
> > using itemTypes it copies rather than filters. For example:
> >
> >     <div ex:role="collection" ex:baseCollectionID="default"
> > ex:itemTypes="Family" id="the-families"></div>
> >     <div ex:role="collection" ex:baseCollectionID="default"
> > ex:itemTypes="Person" id="the-people"></div>
> >     <div ex:role="collection" ex:baseCollectionID="default"
> > ex:itemTypes="Marriage, Birth, Death" id="the-events"></div>
> >
> > Even with baseCollectionID in there, my facets still don't work right.
> > Any view filtered by the above collectionID is not phased by a facet
> > unless that facet has the same collectionID, which is a pain because I
> > wanted my one set of facets to work for all views. If I use
> > ex:expression instead of ex:itemTypes, how do I filter on type? How do
> > I limit to multiple types like that third line? It would be nice if
> > you could put the ex:itemTypes right on your view. Thanks for your time.
> Actually what you need is something like this
>
>     <div ex:role="collection" ex:itemTypes="Person"></div>
>
>     <div ex:role="collection"
>        ex:baseCollectionID="default"
>        ex:expression=".family"></div>
>
> Define the default collection and then tie the other collections to it
> using ex:expression. That's presuming that your data looks something
> like this:
>
>     {   type: "Person",
>         label: "Homer Simpson",
>         family: "The Simpson Family",
>         ...
>     }
>
>     ...
>
>     {   type: "Family",
>          label: "The Simpson Family",
>          ...
>
>     }
>
> David
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