Suppose I have two models:
class A < ActiveRecord::Base has_many :bs end
class B < ActiveRecord::Base belongs_to :a end
Now, if I say in a controller b1 = @a.build(params[:b]) I can build a new B associated to an A.
If the @ is also new, I need to also do this b1.a = @a but if the @a is not new, then its id is filled in for me.
(1) Why? Even though there is no @a.id to set the b1.a to, there is an @a object that I can set b1.a to, and I do. Couldn't build do this for me so @a.build(...) would work in both cases?
Either way, if I now say @a.save! Then not only are the values of @a that may have been updated saved, the new object b1 is also saved to the database. Not only that, this all happens in the same transaction. Yea!
However, instead if I were to update an existing b as follows: b1 = @a.bs[1].update(params[:b]) and then @a.save! then this time, the updated values of b1 are not saved to the database.
(2) Why? Wouldn't it be cleaner to update the associated b values into the database as well during the save of the @a ?
Further, if I want the @a and the associated @a.bs to all be sent to the database in the same transaction, as far as I can tell, I have the following choices.
In my controller save the @a and @a.bs in one transaction: ActiveRecord::Base.transaction do @a.bs.each do |b| b.save! end @a.save! end
Or perhaps to be more elegant I could push all that into a method save_it! on class A.
Or I can try to figure out how to get the @a.bs to save in the same transaction as the @a. I suppose I can do that as follows:
class A < ActiveRecord::Base before_save do |a| bs.each do |b| b.save! end end end
This is a bit better, but -- amazingly -- I have somehow at times gotten into an infinite loop doing this, though it was a bit more complex than that. Even though the documentation says that if I save on the belongs_to side, the object belonged-to will not be saved, I had a case where saving the associated objects like this ended up recursing on itself. Is this idiom supposed to work and was I therefore getting an infinite loop because (a) I was not doing it quite right, or (b) is this idiom not good.
(3) What is the right idiom for this situation, where I want to save the @a and the associated bs in one transaction?
Daniel