Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

PHP

Susitha Herath Mudiyanselage
Susitha Herath Mudiyanselage
4,957 Points
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
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
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
Christopher Davis
Christopher Davis
13,738 Points

You didn't pluralize the genre table name.