Reputation: 9013
I have the following simple DB:
Table Types:
- ID int
- TypeName nvarchar
Table Users:
- ID int
- UserName nvarchar
- TypeID int
Table BusyTime
- ID int
- UserID int
- BTime time(0)
But one restriction - records in BusyTime should be only for users, which have TypeID = 3. Users with TypeID = 1 and with TypeID = 2 can't have records in BusyTime (it contradicts the business logic) How to describe it on MS SQL level? Or should I redesign DB ?
Upvotes: 0
Views: 62
Reputation: 239646
I'm assuming your primary keys in each table are just on ID
. What you need to change is, add a UNIQUE KEY
constraint on both ID
and TypeID
in Users
:
ALTER TABLE Users ADD CONSTRAINT UQ_User_Types_XRef (ID,TypeID)
And create the BusyTime
table as:
CREATE TABLE BusyTime (
ID int not null,
UserID int not null,
BTime time(0) not null,
_Type_XRef as 3 persisted,
constraint PK_BusyTime PRIMARY KEY (ID),
constraint FK_BusyTime_Users FOREIGN KEY (UserID)
references Users (ID),
constraint FK_BusyTime_Users_XRef FOREIGN KEY (UserID,_Type_XRef)
references Users (ID,TypeID)
)
Where I've assumed PK_BusyTime
and FK_BusyTime_Users
were your existing constraints. It's a matter of taste whether you drop FK_BusyTime_Users
(which is the "real" foreign key constraint) now that FK_BusyTime_Users_XRef
exists.
Upvotes: 2
Reputation: 1269543
Setting up such a conditional constraint is possible. One method is to add compound index to users
:
create unique index idx_users(type, id) on users(type, id)
Then use this for a foreign key constraint:
alter table busytime add _type as (3) persisted not null;
alter table busytime add constraint foreign key (_type, userId) on users(type, id);
Unfortunately, I think the persisted
is needed for this column so it actually occupies space in the record.
Unfortunately, I don't think this works either:
alter table busytime add constraint foreign key (3, userId) on users(type, id);
Upvotes: 0
Reputation: 584
You can use check constraints to disallow invalid types: https://technet.microsoft.com/en-us/library/ms188258%28v=sql.105%29.aspx
Upvotes: 0