How can I implement a foreign key in SQLite? I was thinking something like this:
CREATE TABLE job (_id INTEGER PRIMARY KEY AUTOINCREMENT, employer_id INTEGER, ...);
CREATE TABLE employer(_id INTEGER, employer_name TEXT NOT NULL, ...);
Where employer_id is the _id from the table employer. Would this work? Is there another fast, maybe less prone to error way? Maybe with triggers?
Maybe I don't understand the question, but if it's the constraint you want, just do this:
ALTER TABLE Job
ADD FOREIGN KEY (employer_id)
REFERENCES Employer(_id)
ON DELETE NO ACTION
ON UPDATE NO ACTION;
See SQLite (3.6.19) Foreign Key Support
(Earlier version of SQLite do not support enforced FK relationships.)
Related
what'S wrong with the following statement?
ALTER TABLE submittedForecast
ADD CONSTRAINT FOREIGN KEY (data) REFERENCES blobs (id);
The error message I am getting is
Can't create table `fcdemo`.`#sql-664_b` (errno: 150 "Foreign key constraint is incorrectly formed")
This works for me on MariaDB 10.1.8:
CREATE TABLE `submittedforecast` (
`id` INT(11) NOT NULL,
`data` INT(11) NOT NULL,
PRIMARY KEY (`id`),
INDEX `data` (`data`)
) ENGINE=InnoDB;
CREATE TABLE `blobs` (
`id` INT(11) NOT NULL,
`content` BLOB NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB;
ALTER TABLE submittedForecast
ADD CONSTRAINT FOREIGN KEY (data) REFERENCES blobs (id);
Can you give your MariaDB version number and a complete example including the CREATE TABLE statements for submittedForecast and blobs?
No idea if you already solved this but make sure both engines and collations match between tables (e.g: latin1 to latin1 and InnoDB to InnoDB).
I was having the same issue and by switching these I managed to get it working.
I was getting the same issue and after looking at my table structure, found out that my child table's foreign key clause was not referencing the primary key of parent table. Once i changed it to reference to primary key of parent table, the error was gone.
I had the same error and is actually pretty easy to solve, you have name the constraint, something like this should do:
ALTER TABLE submittedForecast ADD CONSTRAINT `fk_submittedForecast`
FOREIGN KEY (data) REFERENCES blobs (id)
If you would like more cohesion also add at the end of the query
ON DELETE CASCADE ON UPDATE RESTRICT
This could also be a different fields error so check if the table key and the foreign key are of the same type and have the same atributes.
I had the same problem, too. When checking the definition of the fields, I noticed that one field was defined as INT and the other as BIGINT. After changing the BIGINT type to INT, I was able to create my foreign key.
Using the C API, I don't see a way to determine the foreign key constraints for a named table?
Given this example:
CREATE TABLE artist(
artistid INTEGER PRIMARY KEY,
artistname TEXT
);
CREATE TABLE track(
trackid INTEGER,
trackname TEXT,
trackartist INTEGER,
FOREIGN KEY(trackartist) REFERENCES artist(artistid)
);
sqlite3_table_column_metadata() will tell you it's a primary key, autoincrement, etc. but how
do I get the foreign key constraints?
FOREIGN KEY(trackartist) REFERENCES artist(artistid)
I want to be able to get a list back for table "track" that there are foreign keys back to table Artist column artistid?
I don't see an api to do this? I need to do this programmaticlly upon opening the database, for purposes of aggregation.
Thanks.
After using PRAGMA foreign_key_list(Valuation);
I got back:
PRAGMA foreign_key_list(Valuation);
0|0|Stock|StockId|Id|NO ACTION|NO ACTION|NONE
I understand I need to split on the vertical bar, but what are the first two columns? 0|0 ?
Please note that (foreign) keys can consist of multiple columns, so it would not make sense to return this as column information.
To get information about a table's foreign keys, use this:
PRAGMA foreign_key_list(table-name);
This pragma returns one row for each foreign key constraint created by a REFERENCES clause in the CREATE TABLE statement of table "table-name".
I want to modify the update option of one foreign key.
For this I executed this command:
alter table testusers.ORDERS
DROP CONSTRAINT ORDER_FK_2,
ADD CONSTRAINT ORDER_FK_2 FOREIGN KEY(FK_PRODUCER_ID) REFERENCES testuser.PRODUCER (producer_id)
ON UPDATE CASCADE ON DELETE CASCADE;
If I execute this, there is the following error:
SQL-Fehler: ORA-01735: Ungültige Option ALTER TABLE
01735. 00000 - "invalid ALTER TABLE option"
There is no comma separated list for the alter table according to documentation syntax diagram http://docs.oracle.com/cd/B28359_01/server.111/b28286/clauses002.htm#CJAEDFIB
create table orders(order_id number, fk_producer_id number, CONSTRAINT order_pk PRIMARY KEY (order_id));
create table producer(producer_id number, CONSTRAINT producer_pk PRIMARY KEY (producer_id));
alter table orders
ADD CONSTRAINT ORDER_FK_2 FOREIGN KEY( FK_PRODUCER_ID)
REFERENCES PRODUCER (producer_id) ;
alter table orders
DROP CONSTRAINT ORDER_FK_2;
alter table orders
ADD CONSTRAINT ORDER_FK_2 FOREIGN KEY( FK_PRODUCER_ID)
REFERENCES PRODUCER (producer_id) ;
Ahm, yes, and I could not find any ON UPDATE CASCADE syntax either. But I am sure you can work it out now. Otherwise drop a little comment or post a new question.
Recently I started using SQLite (as required for my study) and I came accross a couple of restrictions of SQLite and I was wondering: can't SQLite create foreign keys on the same table? E.g. this is my code:
CREATE TABLE Categories
(
name varchar(20),
parent_category varchar(20) NULL,
PRIMARY KEY(name),
FOREIGN KEY parent_category_fk(parent_category) REFERENCES Categories(name)
)
But it gives me an error for the foreign key when I try to execute the SQL in SQLiteStudio.
Does anyone know why this isn't working?
The problem is that you have the wrong syntax for the FK clause. It should be:
FOREIGN KEY (parent_category) REFERENCES Categories(name)
If you want to name the FK constraint, you do that with a prefix of the CONSTRAINT keyword, like this:
CONSTRAINT parent_category_fk FOREIGN KEY (parent_category) REFERENCES Categories(name)
I know that SQLite does not enforce foreign keys natively, but that's not my primary concern. The question is: If I declare
CREATE TABLE invoice (
invoiceID INTEGER PRIMARY KEY,
clientID INTEGER REFERENCES client(clientID),
...
)
will sqlite at least use the information that clientID is a foreign key to optimize queries and automatically index invoice.clientID, or is this constraint a real no-op?
In the SQLite Documentation it says:
... "an index should be created on the child key columns of each foreign key constraint"
ie. the index is not automatically created, but you should create one in every instance.
Even if it is not actually a no-op (a data structure describing the constraint is added to the table), foreign key related statement doesn't create any index on involved columns.
Indexes are implicitly created only in the case of PRIMARY KEY and UNIQUE statements.
For more details, check it out build.c module on the sqlite source tree:
http://www.sqlite.org/cvstrac/rlog?f=sqlite/src/build.c https://www.sqlite.org/src/file?name=src/build.c&ci=tip