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PHP

Posted by Susitha Herath Mudiyanselage
Susitha Herath Mudiyanselage
4,957 Points

Help!

Database challenge is not passing!

Alter the "t_movies" table to add a foreign key called "fk_genre_id" and constrain it to reference the "t_genres" "pk_id".

my answer: ALTER TABLE t_movies ADD fk_genre_id, ADD CONSTRAINT FORIEGN KEY (fk_genre_id) REFERENCES t_genres(pk_id)

1 Answer

Christopher Davis
Christopher Davis
13,738 Points

Is that your code copied verbatim? If so, you have a misspelling: "FORIEGN KEY" should be ''FOREIGN KEY''

Susitha Herath Mudiyanselage
Susitha Herath Mudiyanselage
4,957 Points

I have corrected the spellings and done it again with ALTER TABLE t_movies ADD COLUMN fk_genre_id INTEGER NULL, ADD CONSTRAINT FOREIGN KEY(fk_genre_id) REFERENCES t_genre (pk_id);

I get these errors..

You are missing the fk_genre_id column!

and

SQL Error: Can't create table 'mysql_68e56cb3_c6e4_4d25_aace_8b798c81ad5f_cc_db.#sql-6d7a_34f4' (errno: 150)

Christopher Davis
Christopher Davis
13,738 Points

You didn't pluralize the genre table name.

Susitha Herath Mudiyanselage
Susitha Herath Mudiyanselage
4,957 Points

It Worked! Thank you very much for your reply.