[Ruby-on-rails] delete_all vs destroy_all?


delete_all is a single SQL DELETE statement and nothing more. destroy_all calls destroy() on all matching results of :conditions (if you have one) which could be at least NUM_OF_RESULTS SQL statements.

If you have to do something drastic such as destroy_all() on large dataset, I would probably not do it from the app and handle it manually with care. If the dataset is small enough, you wouldn't hurt as much.


I am looking for the best approach to delete records from a table. For instance, I have a user whose user ID is across many tables. I want to delete this user and every record that has his ID in all tables.

u = User.find_by_name('JohnBoy')

This works and removes all references of the user from all tables, but I heard that destroy_all was very process heavy, so I tried delete_all. It only removes the user from his own user table and the id from all the other tables are made null, but leaves the records intact in them. Can someone share what the correct process is for performing a task like this?

I see that destroy_all calls the destroy function on all associated objects but I just want to confirm the correct approach.

I’ve made a small gem that can alleviate the need to manually delete associated records in some circumstances.

This gem adds a new option for ActiveRecord associations:

dependent: :delete_recursively

When you destroy a record, all records that are associated using this option will be deleted recursively (i.e. across models), without instantiating any of them.

Note that, just like dependent: :delete or dependent: :delete_all, this new option does not trigger the around/before/after_destroy callbacks of the dependent records.

However, it is possible to have dependent: :destroy associations anywhere within a chain of models that are otherwise associated with dependent: :delete_recursively. The :destroy option will work normally anywhere up or down the line, instantiating and destroying all relevant records and thus also triggering their callbacks.