Created
May 16, 2011 02:18
-
-
Save DGaffney/973809 to your computer and use it in GitHub Desktop.
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
| #Table Friendships | |
| +----+------------+-------------------+------------------+------------+------------+ | |
| | id | dataset_id | following_user_id | followed_user_id | created_at | deleted_at | | |
| +----+------------+-------------------+------------------+------------+------------+ | |
| | 1 | 1 | 12 | 13348 | NULL | NULL | | |
| | 2 | 1 | 107 | 13348 | NULL | NULL | | |
| | 3 | 1 | 274 | 13348 | NULL | NULL | | |
| | 4 | 1 | 324 | 13348 | NULL | NULL | | |
| | 5 | 1 | 418 | 13348 | NULL | NULL | | |
| | 6 | 1 | 422 | 13348 | NULL | NULL | | |
| | 7 | 1 | 455 | 13348 | NULL | NULL | | |
| | 8 | 1 | 456 | 13348 | NULL | NULL | | |
| | 9 | 1 | 509 | 13348 | NULL | NULL | | |
| | 10 | 1 | 544 | 13348 | NULL | NULL | | |
| | 11 | 1 | 12 | 13349 | NULL | NULL | | |
| | 12 | 1 | 107 | 13349 | NULL | NULL | | |
| | 13 | 1 | 274 | 13349 | NULL | NULL | | |
| | 14 | 1 | 324 | 13349 | NULL | NULL | | |
| | 15 | 1 | 418 | 13349 | NULL | NULL | | |
| | 16 | 1 | 422 | 13349 | NULL | NULL | | |
| | 17 | 1 | 455 | 13350 | NULL | NULL | | |
| | 18 | 1 | 456 | 13350 | NULL | NULL | | |
| | 19 | 1 | 509 | 13350 | NULL | NULL | | |
| | 20 | 1 | 544 | 13350 | NULL | NULL | | |
| +----+------------+-------------------+------------------+------------+------------+ | |
| #With data in a structure like this, how would I be able to find out the intersections of following_user_id from followed_user_id? In other words, followed_user_id of | |
| #13348 and 13349 have 6 similar start_id rows (following_user_ids of 12,107,274,324,418,422), 13348 and 13350 have 4 (455,456,509,544), and 13349 | |
| #and 13350 have zero similar rows. | |
| #Query I have been trying: select count(a.following_user_id) from friendships as a, friendships as b, friendships as c where a.following_user_id=b.following_user_id=c.following_user_id and (a.followed_user_id=13348 and b.followed_user_id=13349 and c.followed_user_id=13350); |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment