The FOREIGN KEY constraint is a key used to link two tables together.
You could simply create two columns in Ticket, OwnedByUserId and OwnedByGroupId, and have nullable Foreign Keys to each table. You could create M:M reference tables enabling both ticket:user and ticket:group relationships. Perhaps in future you will want to allow a single ticket to be owned by multiple users or groups?
MySQL allows us to add a FOREIGN KEY constraint on multiple columns in a table. The condition is that each Foreign Key in the child table must refer to the different parent table.
Note that foreign keys are not mandatory, and a table may have no foreign keys. Conversely, every column in a table may have a foreign key constraint.
You have a few options, all varying in "correctness" and ease of use. As always, the right design depends on your needs.
You could simply create two columns in Ticket, OwnedByUserId and OwnedByGroupId, and have nullable Foreign Keys to each table.
You could create M:M reference tables enabling both ticket:user and ticket:group relationships. Perhaps in future you will want to allow a single ticket to be owned by multiple users or groups? This design does not enforce that a ticket must be owned by a single entity only.
You could create a default group for every user and have tickets simply owned by either a true Group or a User's default Group.
Or (my choice) model an entity that acts as a base for both Users and Groups, and have tickets owned by that entity.
Heres a rough example using your posted schema:
create table dbo.PartyType
(
PartyTypeId tinyint primary key,
PartyTypeName varchar(10)
)
insert into dbo.PartyType
values(1, 'User'), (2, 'Group');
create table dbo.Party
(
PartyId int identity(1,1) primary key,
PartyTypeId tinyint references dbo.PartyType(PartyTypeId),
unique (PartyId, PartyTypeId)
)
CREATE TABLE dbo.[Group]
(
ID int primary key,
Name varchar(50) NOT NULL,
PartyTypeId as cast(2 as tinyint) persisted,
foreign key (ID, PartyTypeId) references Party(PartyId, PartyTypeID)
)
CREATE TABLE dbo.[User]
(
ID int primary key,
Name varchar(50) NOT NULL,
PartyTypeId as cast(1 as tinyint) persisted,
foreign key (ID, PartyTypeId) references Party(PartyID, PartyTypeID)
)
CREATE TABLE dbo.Ticket
(
ID int primary key,
[Owner] int NOT NULL references dbo.Party(PartyId),
[Subject] varchar(50) NULL
)
The first option in @Nathan Skerl's list is what was implemented in a project I once worked with, where a similar relationship was established between three tables. (One of them referenced two others, one at a time.)
So, the referencing table had two foreign key columns, and also it had a constraint to guarantee that exactly one table (not both, not neither) was referenced by a single row.
Here's how it could look when applied to your tables:
CREATE TABLE dbo.[Group]
(
ID int NOT NULL CONSTRAINT PK_Group PRIMARY KEY,
Name varchar(50) NOT NULL
);
CREATE TABLE dbo.[User]
(
ID int NOT NULL CONSTRAINT PK_User PRIMARY KEY,
Name varchar(50) NOT NULL
);
CREATE TABLE dbo.Ticket
(
ID int NOT NULL CONSTRAINT PK_Ticket PRIMARY KEY,
OwnerGroup int NULL
CONSTRAINT FK_Ticket_Group FOREIGN KEY REFERENCES dbo.[Group] (ID),
OwnerUser int NULL
CONSTRAINT FK_Ticket_User FOREIGN KEY REFERENCES dbo.[User] (ID),
Subject varchar(50) NULL,
CONSTRAINT CK_Ticket_GroupUser CHECK (
CASE WHEN OwnerGroup IS NULL THEN 0 ELSE 1 END +
CASE WHEN OwnerUser IS NULL THEN 0 ELSE 1 END = 1
)
);
As you can see, the Ticket
table has two columns, OwnerGroup
and OwnerUser
, both of which are nullable foreign keys. (The respective columns in the other two tables are made primary keys accordingly.) The CK_Ticket_GroupUser
check constraint ensures that only one of the two foreign key columns contains a reference (the other being NULL, that's why both have to be nullable).
(The primary key on Ticket.ID
is not necessary for this particular implementation, but it definitely wouldn't harm to have one in a table like this.)
Yet another option is to have, in Ticket
, one column specifying the owning entity type (User
or Group
), second column with referenced User
or Group
id and NOT to use Foreign Keys but instead rely on a Trigger to enforce referential integrity.
Two advantages I see here over Nathan's excellent model (above):
Another approach is to create an association table that contains columns for each potential resource type. In your example, each of the two existing owner types has their own table (which means you have something to reference). If this will always be the case you can have something like this:
CREATE TABLE dbo.Group
(
ID int NOT NULL,
Name varchar(50) NOT NULL
)
CREATE TABLE dbo.User
(
ID int NOT NULL,
Name varchar(50) NOT NULL
)
CREATE TABLE dbo.Ticket
(
ID int NOT NULL,
Owner_ID int NOT NULL,
Subject varchar(50) NULL
)
CREATE TABLE dbo.Owner
(
ID int NOT NULL,
User_ID int NULL,
Group_ID int NULL,
{{AdditionalEntity_ID}} int NOT NULL
)
With this solution, you would continue to add new columns as you add new entities to the database and you would delete and recreate the foreign key constraint pattern shown by @Nathan Skerl. This solution is very similar to @Nathan Skerl but looks different (up to preference).
If you are not going to have a new Table for each new Owner type then maybe it would be good to include an owner_type instead of a foreign key column for each potential Owner:
CREATE TABLE dbo.Group
(
ID int NOT NULL,
Name varchar(50) NOT NULL
)
CREATE TABLE dbo.User
(
ID int NOT NULL,
Name varchar(50) NOT NULL
)
CREATE TABLE dbo.Ticket
(
ID int NOT NULL,
Owner_ID int NOT NULL,
Owner_Type string NOT NULL, -- In our example, this would be "User" or "Group"
Subject varchar(50) NULL
)
With the above method, you could add as many Owner Types as you want. Owner_ID would not have a foreign key constraint but would be used as a reference to the other tables. The downside is that you would have to look at the table to see what the owner types there are since it isn't immediately obvious based upon the schema. I would only suggest this if you don't know the owner types beforehand and they won't be linking to other tables. If you do know the owner types beforehand, I would go with a solution like @Nathan Skerl.
Sorry if I got some SQL wrong, I just threw this together.
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