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Richard Stringfellow
6,883 PointsJoining Relational Data Between Tables in SQL
Quiz question 2 at the end of Stage 6, "Joining Relational Data Between Tables in SQL has me stumped. I seem to get a better result, though my answer is still not correct, if I use "movies" instead of "movie", and "genres" instead of "genre". Here are some of the things I have tried, and the results I have gotten.
SELECT movie.title, genre.name FROM movie LEFT OUTER JOIN genre ON genre.name; (returns) Bummer! There's something wrong with your SQL statement. Please review your code and try again.
(plus 11 variations and permutations of the above, all yielding the same result)
SELECT movies.title, genres.name FROM movies LEFT OUTER JOIN genres ON genres.name; (returns) Bummer! You're not retrieving the movie 'title' first and the genre 'name' second with all information from the 'movies' table. Use an OUTER JOIN. Use LEFT if the movies table is on the left or RIGHT if it's on the right of the statement.
4 Answers
Justin Horner
Treehouse Guest TeacherHello Richard,
You need to join on the id column of the genre table and the genre_id column of the movies table.
SELECT g.name, m.title FROM movies m LEFT OUTER JOIN genres g ON m.genre_id = g.id
I hope this helps.
Richard Stringfellow
6,883 PointsYes, it did help. I feel like the function of the ON clause was not adequately explained in the course.
David Jones
1,635 PointsJustin can you explain the answer a bit more? Ive tried to do the same but it doesnt work (although in MYsql itslef it sort of works but then tells me that Aliens is a musical lol)
SELECT genres.name, movies.title FROM movies LEFT OUTER JOIN genres ON movies.genre_id = genre_id;
I left out the 'm' and 'g' that you used in your example as I dont understand why they are being used and also they dont work too.
Adam Oliver
8,214 PointsIts very similar to the question 1 answer look-> SELECT movies.title, genres.name FROM movies LEFT OUTER JOIN genres ON movies.genre_id = genres.id;