Re: ALTER TABLE to add Foreign Key Constraint

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От Robert B. Easter
Тема Re: ALTER TABLE to add Foreign Key Constraint
Дата
Msg-id 00061723004000.18758@comptechnews
обсуждение исходный текст
Ответ на ALTER TABLE to add Foreign Key Constraint  (Vipin Samtani <vipin@attglobal.net>)
Список pgsql-general
On Sat, 17 Jun 2000, Vipin Samtani wrote:
> I am trying to designate foreign keys after tables "distributors" and
> "addresses" have already been created with primary keys "distributor"
> and "address" respectively. I am attempting to use ALTER TABLE to
> accomplish this. I found this statement in the documentation
>
> ALTER TABLE distributors ADD CONSTRAINT distfk FOREIGN KEY (address)
> REFERENCES addresses(address) MATCH FULL
>
>
> When I run it, I get this error:
>
> ERROR:  parser: parse error at or near "constraint"
>
> What is wrong with the statement?

Consider this example:

-- Load the PGSQL procedural language
-- This could also be done with the createlang script/program.
-- See man createlang.
CREATE FUNCTION plpgsql_call_handler()
    RETURNS OPAQUE AS '/usr/local/pgsql/lib/plpgsql.so'
    LANGUAGE 'C';
CREATE TRUSTED PROCEDURAL LANGUAGE 'plpgsql'
    HANDLER plpgsql_call_handler
    LANCOMPILER 'PL/pgSQL';

CREATE TABLE test1 (id INTEGER PRIMARY KEY);
CREATE TABLE test2 (id INTEGER PRIMARY KEY);

ALTER TABLE test1 ADD CONSTRAINT test1_id_fk
    FOREIGN KEY (id) REFERENCES test2
    ON UPDATE CASCADE
    ON DELETE CASCADE -- can delete test2
    INITIALLY DEFERRED;
ALTER TABLE test2 ADD CONSTRAINT test1_id_fk
    FOREIGN KEY (id) REFERENCES test1
    ON UPDATE CASCADE
    ON DELETE RESTRICT -- disallows delete test1
    INITIALLY DEFERRED;

CREATE SEQUENCE test_id_seq;

CREATE FUNCTION new_tests() RETURNS INTEGER AS '
DECLARE
    new_seq INTEGER;
BEGIN
    new_seq := nextval(''test_id_seq'');
    INSERT INTO test1 VALUES (new_seq);
    INSERT INTO test2 VALUES (new_seq);
    RETURN new_seq;
END;
' LANGUAGE 'plpgsql';

-- implicit BEGIN;
SELECT new_tests();
-- implicit COMMIT;
SELECT new_tests();
SELECT new_tests();

SELECT * FROM test1;
SELECT * FROM test2;

DELETE FROM test1 WHERE id = 1; -- this will fail
DELETE FROM test2 WHERE id = 1; -- this will succeed and cacade

SELECT * FROM test1;
SELECT * FROM test2;

--
Robert B. Easter

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