我有一张由人们提交的视频表:
+----+-----------+------+---------+
| id | by_person | type | title |
+----+-----------+------+---------+
| 1 | 3 | 1 | title1 |
| 2 | 4 | 1 | title2 |
| 3 | 3 | 1 | title3 |
| 4 | 4 | 2 | title4 |
| 5 | 3 | 1 | title5 |
| 6 | 6 | 2 | title6 |
| 7 | 6 | 2 | title7 |
| 8 | 4 | 2 | title8 |
| 9 | 3 | 1 | title9 |
| 10 | 4 | 1 | title10 |
| 11 | 4 | 1 | title11 |
| 12 | 3 | 1 | title12 |
+----+-----------+------+---------+
如何SELECT
提交最多 type=1 视频的前两名人员,以便我得到这样的演示文稿?
1. Person(3) - 5 videos
2. Person(4) - 3 videos
最佳答案
我认为这会起作用:
SELECT by_person, count(*) AS total
FROM videos
WHERE type = 1
GROUP BY by_person
ORDER BY total DESC
LIMIT 2
关于php - MySQL 选择提交视频最多的前两个人,其中 type=1,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16084444/