Rails migration create_join_table with uuids

Question

Newer versions of rails let you specify that tables should be created with a uuid primary key like so:

create_table :foos, id: :uuid do |t|
  # ...
end

Which is great. And for a long time rails has supported creating join tables like so:

create_join_table :foos, :bars do |t|
  # ...
end

Also great. Except my tables have uuid primary keys and that generates foreign key columns of type integer instead of type uuid.

Looking over the documentation for create_join_table, I can't find anything obvious to change the column type. Is it possible to use create_join_table with uuids?

Or do I have create the join table manually:

create_table :bars_foos, id: false do |t|
  t.uuid :bar_id
  t.uuid :foo_id
end

anka · Accepted Answer · 2017-06-13 13:47:58Z

13

Within Rails 5.0 you can use an additional option column_options on the create_join_table method to specify the type of your id columns. Your migration would then look like:

create_join_table :foos, :bars, column_options: {type: :uuid} do |t|
  t.index [:foo_id, :baar_id]
end

answered Jun 13, 2017 at 13:47

anka

3,8571 gold badge32 silver badges38 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

cabgfx Over a year ago

Confirmed working with Rails 7 👍

ImaginateWayne · Accepted Answer · 2016-07-11 06:19:32Z

9

There is no way of creating join tables with uuids.

As pointed out in the question create_table is the only option. The best way of emulating create_join_tables with uuid is by using create_tables as follows:

Run: rails g migration CreateFoosBars bars:references foos:references
The command will produce the following output which you will need to modify

generate output

class CreateBarFoos < ActiveRecord::Migration
  def change
    create_table :bars_foos, id: :uuid do |t|
      t.references :bars, foreign_key: true
      t.references :foo, foreign_key: true
    end
  end
end

Change id: uuid => id: false
Add type: uuid, index: true to the end of the references

final migration

class CreateBarFoos < ActiveRecord::Migration
  def change
    create_table :bars_foos, id: false do |t|
      t.references :bars, foreign_key: true, type: :uuid, index: true
      t.references :foo, foreign_key: true, type: :uuid, index: true
    end
  end
end

It would be good if Rails could add extra support for different id types in create_join_table, this could even be inferred by an existing migration.

Until then hopefully these steps will achieve the same result.

edited Jul 11, 2016 at 6:19

answered Jul 11, 2016 at 6:06

ImaginateWayne

5301 gold badge5 silver badges14 bronze badges

1 Comment

Todd Baur Over a year ago

This is the right answer for the given version of Rails (4.2.x), and @anka's answer is correct for version 5 and up.

ihaztehcodez · Accepted Answer · 2015-10-02 23:34:22Z

5

I should have looked at the code...

def create_join_table(table_1, table_2, options = {})
  join_table_name = find_join_table_name(table_1, table_2, options)

  column_options = options.delete(:column_options) || {}
  column_options.reverse_merge!(null: false)

  t1_column, t2_column = [table_1, table_2].map{ |t| t.to_s.singularize.foreign_key }

  create_table(join_table_name, options.merge!(id: false)) do |td|
    td.integer t1_column, column_options
    td.integer t2_column, column_options
    yield td if block_given?
  end
end

Columns are explicitly created as integers with no means to change them. Too bad...

answered Oct 2, 2015 at 23:34

ihaztehcodez

2,14317 silver badges29 bronze badges

Collectives™ on Stack Overflow

Rails migration create_join_table with uuids

3 Answers 3

1 Comment

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related