Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

factory_girl association validations

I have an issue where I have a parent model Foo, which both has_many :bars and has_many :bazes. Finally, I also have a join model BarBaz which belongs_to :bar and belongs_to :baz. I want to validate all bar_bazes so that its bar and baz both belong to the same foo. But I can't seem to figure out a way to define a factory for this model that would be valid.

Factory.define(:bar) do |bar|
  bar.association(:foo)
end

Factory.define(:baz) do |baz|
  bar.association(:foo)
end

Factory.define(:bar_baz) do |bar_baz|
  baz_bar.association(:foo)
  baz_bar.association(:bar)
  baz_bar.association(:baz)
end

I get an invalid record error when I try to create the latter, because the bar and baz factory_girl tries to associate it each have their own foo. Am I screwed?

like image 775
tfwright Avatar asked Jan 13 '10 02:01

tfwright


1 Answers

So after hours of beating my head against this issue, I think I finally have a solution. It's pretty insane though, so hopefully someone else can still show me where I'm being stupid.

Factory.define :foo do |foo|
end

Factory.define :bar do |bar|
end

Factory.define :baz do |baz|
end

Factory.define :foo_with_baz do |foo|
  foo.after_create { |foo| Factory(:baz, :foo => foo) }
end

Factory.define :bar_baz do |bar_baz|
  bar_baz.bar {|bar| bar.association(:bar, :foo => Factory(:foo_with_baz))
  bar_baz.after_build {| bar_baz| bar_baz.baz_id = bar_baz.foo.bars.first.id }
end

The key issue being that there needs to be a foo in the database already you can get at through the factories alone, since you can use local variables or arbitrary ruby code in factories.rb (as far as I can tell).

like image 124
tfwright Avatar answered Sep 29 '22 10:09

tfwright