Create unique constraint with null columns

2019-01-01 08:32发布

问题:

I have a table with this layout:

CREATE TABLE Favorites
(
  FavoriteId uuid NOT NULL PRIMARY KEY,
  UserId uuid NOT NULL,
  RecipeId uuid NOT NULL,
  MenuId uuid
)

I want to create a unique constraint similar to this:

ALTER TABLE Favorites
ADD CONSTRAINT Favorites_UniqueFavorite UNIQUE(UserId, MenuId, RecipeId);

However, this will allow multiple rows with the same (UserId, RecipeId), if MenuId IS NULL. I want to allow NULL in MenuId to store a favorite that has no associated menu, but I only want at most one of these rows per user/recipe pair.

The ideas I have so far are:

  1. Use some hard-coded UUID (such as all zeros) instead of null.
    However, MenuId has a FK constraint on each user\'s menus, so I\'d then have to create a special \"null\" menu for every user which is a hassle.

  2. Check for existence of a null entry using a trigger instead.
    I think this is a hassle and I like avoiding triggers wherever possible. Plus, I don\'t trust them to guarantee my data is never in a bad state.

  3. Just forget about it and check for the previous existence of a null entry in the middle-ware or in a insert function, and don\'t have this constraint.

I\'m using Postgres 9.0.

Is there any method I\'m overlooking?

回答1:

Create two partial indexes:

CREATE UNIQUE INDEX favo_3col_uni_idx ON favorites (user_id, menu_id, recipe_id)
WHERE menu_id IS NOT NULL;

CREATE UNIQUE INDEX favo_2col_uni_idx ON favorites (user_id, recipe_id)
WHERE menu_id IS NULL;

This way, there can only be one combination of (user_id, recipe_id) where menu_id IS NULL, effectively implementing the desired constraint.

Possible drawbacks: you cannot have a foreign key referencing (user_id, menu_id, recipe_id) this way, you cannot base CLUSTER on a partial index, and queries without a matching WHERE condition cannot use the partial index.

It seems unlikely you\'d want a FK reference three columns wide (use the PK column instead). If you need a complete index, you can alternatively drop the WHERE condition from favo_3col_uni_idx and your requirements are still enforced.
The index, now comprising the whole table, overlaps with the other one and gets bigger. Depending on typical queries and the percentage of NULL values, this may or may not be useful. In extreme situations it might even help to maintain both versions of favo_3col_uni_idx.

Aside: I advise not to use mixed case identifiers in PostgreSQL.



回答2:

You could create a unique index with a coalesce on the MenuId:

CREATE UNIQUE INDEX
Favorites_UniqueFavorite ON Favorites
(UserId, COALESCE(MenuId, \'00000000-0000-0000-0000-000000000000\'), RecipeId);

You\'d just need to pick a UUID for the COALESCE that will never occur in \"real life\". You\'d probably never see a zero UUID in real life but you could add a CHECK constraint if you are paranoid (and since they really are out to get you...):

alter table Favorites
add constraint check
(MenuId <> \'00000000-0000-0000-0000-000000000000\')


回答3:

You can store favourites with no associated menu in a separate table:

CREATE TABLE FavoriteWithoutMenu
(
  FavoriteWithoutMenuId uuid NOT NULL, --Primary key
  UserId uuid NOT NULL,
  RecipeId uuid NOT NULL,
  UNIQUE KEY (UserId, RecipeId)
)


回答4:

I think there is a semantic problem here. In my view, a user can have a (but only one) favourite recipe to prepare a specific menu. (The OP has menu and recipe mixed up; if I am wrong: please interchange MenuId and RecipeId below) That implies that {user,menu} should be a unique key in this table. And it should point to exactly one recipe. If the user has no favourite recipe for this specific menu no row should exist for this {user,menu} key pair. Also: the surrogate key (FaVouRiteId) is superfluous: composite primary keys are perfectly valid for relational-mapping tables.

That would lead to the reduced table definition:

CREATE TABLE Favorites
( UserId uuid NOT NULL REFERENCES users(id)
, MenuId uuid NOT NULL REFERENCES menus(id)
, RecipeId uuid NOT NULL REFERENCES recipes(id)
, PRIMARY KEY (UserId, MenuId)
);