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PHP

Ambiguous Column

No matter what I come up with, I keep getting 'ambiguous' error messages when I submit this code. I have modified the 'Where' clause every which way I can think of, and it's returning 'main.media.media_id' errors.

SELECT * FROM Media

JOIN Genres ON Genres.genre_id = Media_Genres.genre_id

LEFT OUTER JOIN Media ON Media.media_id = Media_Genres.media_id

WHERE Media.media_id=3;

edit: https://teamtreehouse.com/library/integrating-php-with-databases/using-relational-tables/joining-tables

2 Answers

Simon Coates
Simon Coates
28,694 Points

Can you post the URL? I think you should join media with Media Genres, then join to Genres. As is, i think you're attempting to join media to genres directly (while subsequently making reference to a junction table), while attempting to join that to media again. So if i had to guess you probably want something like:

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

Oh no! I thought for sure this attached to the question but I added the link. Your answer worked perfectly, thank you - I'm still trying to figure out what I did wrong but I'm having a little trouble with the idea of joining databases, so I will have to do some more reading.