I challenge you....

Posted: Tue Apr 16, 2002 3:03 pm (22 years ago)

....to save me from MySQL hell.

On my Fancy Rats site we have a list of name suggestions for people to call their rats which have been contributed by other users of the site. I am making a new database driven version and I want to be able to show the Top 5 name contributors. Below are the relevant parts of the tables I have created and also the result the query should give (I know there aren't 5 but the principal is the same).

Any chance anyone could help me with the MySQL query? I always manage to really confuse myself when more than one table is involved. Let me know if you need any more info.

Code:

ratnames_names
+---------+-----------+-------------+
| name_id | name | contributor |
+---------+-----------+-------------+
| 1 | Ratty | 2 |
| 2 | Splinter | 1 |
| 3 | Tails | 2 |
| 4 | Scruffy | 2 |
| 5 | Rabies | 3 |
| 6 | Hairy | 1 |
+---------+-----------+-------------+

ratnames_contributors
+----------------+------------------+
| contributor_id | contributor_name |
+----------------+------------------+
| 1 | Tom |
| 2 | Dick |
| 3 | Harry |
+----------------+------------------+

Quote:

Result
A. Dick (3 names)
B. Tom (2 names)
C. Harry (1 names)

Posted: Wed Apr 17, 2002 7:30 am (22 years ago)

Don't worry, I got help from else where. Problem is solved. Wink