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Databases

Dylan Van Es
Dylan Van Es
11,085 Points
Posted by Dylan Van Es
Dylan Van Es
11,085 Points

I'm sure I'm on the right track but I just can't get passed this step.

Here's my answer:

SELECT * FROM Media 
JOIN Media_Genres ON Media.genre_id = Media_Genres.genre_id
LEFT OUTER JOIN Genres ON Media.genre_id = Genres.genre_id
WHERE Media.media_id = 3;

2 Answers

Steven Parker
Steven Parker
231,271 Points
Steven Parker
Steven Parker
231,271 Points

Here's a guess.

Instead of "Genres.genre_id", perhaps it should be "Genres.Id"?

If that's not it, please provide a link to the course you are working with and I can give a more accurate answer.

UPDATE — after seeing the instructions:

You have some issues with the joining criteria.

  • the first join should be using the media_id fields instead of genre_id
  • the second join should be ON the Media_Genres.genre_id, there is no "Media.genre_id"
  • while not an error, you don't need to make the second join an OUTER join
Dylan Van Es
Dylan Van Es
11,085 Points
Dylan Van Es
Dylan Van Es
11,085 Points

That worked! Thanks heaps for your help!

Dylan Van Es
Dylan Van Es
11,085 Points
Dylan Van Es
Dylan Van Es
11,085 Points

Thanks Steve,

Unfortunately "genre_id" is the name of the column needed. Here's the whole question:

We will be writing ONLY the SQL query for this challenge.

The library database contains a Media table with the columns media_id, title, img, format, year and category. It also contains a Genres table with the columns genre_id and genre. To join these tables, there is a Media_Genres table that contains the column media_id and genre_id.

Add to the following SELECT statement to JOIN the Media table and the Genres table using the joining table Media_Genres.

SELECT * FROM Media WHERE media_id=3;

NOTE: You will need to add the table to the WHERE clause so that the media_id column is not ambiguous.

The error I'm receiving is "You need to specify tables ON the JOIN." Unfortunately I have no idea what means :-(