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Start your free trialjames white
78,399 PointsSome hints to make up for some limited/bad diagnostic messages in the challenge answer engine..
If any of you have gotten this far (and are still struggling) let me advise you that
some of the Bummer! messages that say something about your "count" being wrong
or unhelpful messages saying error near such and such
will probably not be as helpful in getting to the right answer as
re-watching the video that immediately precedes the challenge question.
This SQL Reporting course (unlike many at Treehouse) does really try to provide
a reasonably tight "coupling" between the SQL code presented in the videos
and what SQL code is expected in the challenges (as least better than using google
to search for relevant SQL code --since their are many different SQL language variations possible).
This course tends toward using SQLite mostly,
but the cheat sheets it provides (in lieu of not having a downloadable zip file for the course)
do include links to multiple types/flavors of SQL:
https://github.com/treehouse/cheatsheets/blob/master/sql_basics/cheatsheet.md
https://github.com/treehouse/cheatsheets/blob/master/reporting_with_sql/cheatsheet.md
A few selected "hints" for the SQL code of various challenges in this aggregate and number functions section:
Link to challenge:
https://teamtreehouse.com/library/reporting-with-sql/aggregate-and-numeric-functions/counting-groups
SELECT genre, COUNT(*) AS genre_count FROM books GROUP BY genre;
SELECT COUNT(DISTINCT genre) AS total_genres FROM books;
Link to challenge:
SELECT MIN(rating) AS star_min, MAX(rating) AS star_max FROM reviews WHERE movie_id = 6;
Link to challenge:
https://teamtreehouse.com/library/reporting-with-sql/aggregate-and-numeric-functions/performing-math
SELECT name, ROUND(price/1.4,2) AS price_gbp FROM products;
Link to challenge:
SELECT AVG(rating) AS average_rating FROM reviews WHERE movie_id = 6;
SELECT SUM(rating) AS starman_total_ratings FROM reviews WHERE movie_id = 6;
More hints for the next (data and time) section can be found here:
A X
12,842 PointsJames thanks so much for the information.
durul
62,690 Pointsdurul
62,690 PointsCounting group challenge answer below.
Challenge Task 1 of 2 <br/> select genre, COUNT(*) as genre_count from books GROUP BY genre; <br> Challenge Task 2 of 2 <br> select COUNT(DISTINCT genre) as total_genres from books;