I have a class like this:
class Foo {
static hasMany = [bars: Bar]
}
When I write:
Foo.getAll()
I get a list of Foo
objects like this:
[ Foo1, Foo2, Foo3 ]
When I write:
Foo.getAll().bars
I get a list of lists of Bar
object like this:
[ [ Bar1, Bar2 ], [ Bar2, Bar3 ], [ Bar1, Bar4 ] ]
But what I want is a unique list of Bar
objects like this:
[ Bar1, Bar2, Bar3, Bar4 ]
My end goal is to have a unique list of ids of the Bar
object in the list above, like this:
[ 1, 2, 3, 4 ]
I have tried variations of the collect method and I have also tried the spread operator but I'm not having any luck.
For generic groovy classes (i.e. non-GORM classes), David is correct that using flatten
and unique
is the best way to go.
In your example though, it looks like you're using GORM domain objects in a many-to-many relationship (otherwise you wouldn't need the uniqueness constraint).
For a domain class, you're best off using either HQL or a criteria to do this in one step. An additional advantage is that the SQL generated for it is much more efficient.
Here's the HQL for getting the unique Bar
ids that are related to any Foo
in a many-to-many relationship:
Bar.executeQuery("select distinct b.id from Foo f join f.bars b")
The criteria for this would look like:
Foo.withCriteria {
bars {
projections {
distinct("id")
}
}
}
One "issue" with using this kind of method is that HQL is not supported in unit tests (and likely will never be) and Criteria queries with join table projections are broken in 2.0.4 unit tests. So any tests around this code would either need to be mocked or use an integration test.
There is a Collection#collectMany
method that works the same way as calling collect
and then flatten
. To get the list of Bar IDs, you might do:
def barIds = Foo.getAll().collectMany { it.bars*.id }
The { it.bars*.id }
closure returns a list with the IDs of the Bars of a Foo (i love these placeholder names hehe), and then collectMany
concatenates those lists into a single list. If you need the IDs to be unique (maybe a Bar can belong to more than one Foo), you can add a .unique()
to that expression :)
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