我有一个这样的表:
mysql> select * from studentscore;
+------------+-----------+-------+
| student_id | cource_id | score |
+------------+-----------+-------+
| 1 | 1 | 80 |
| 1 | 2 | 90 |
| 1 | 3 | 85 |
| 1 | 4 | 78 |
| 2 | 2 | 53 |
| 2 | 3 | 77 |
| 2 | 5 | 80 |
| 3 | 1 | 71 |
| 3 | 2 | 70 |
| 3 | 4 | 80 |
| 3 | 5 | 65 |
| 3 | 6 | 75 |
| 4 | 2 | 90 |
| 4 | 3 | 80 |
| 4 | 4 | 70 |
| 4 | 6 | 95 |
| 5 | 1 | 60 |
| 5 | 2 | 70 |
| 5 | 5 | 80 |
| 5 | 6 | 69 |
| 6 | 1 | 76 |
| 6 | 2 | 88 |
| 6 | 3 | 87 |
| 7 | 4 | 80 |
| 8 | 2 | 71 |
| 8 | 3 | 58 |
| 8 | 5 | 68 |
| 9 | 2 | 88 |
| 10 | 1 | 77 |
| 10 | 2 | 76 |
| 10 | 3 | 80 |
| 10 | 4 | 85 |
| 10 | 5 | 83 |
| 11 | 3 | 80 |
| 12 | 4 | 99 |
| 13 | 5 | 74 |
+------------+-----------+-------+
我想显示 student_id 和高于 80 的学生的平均分数。
我想要的输出是这样的:
+------------+-------------------+
| student_id | Average |
+------------+-------------------+
| 1 | 83.25 |
| 4 | 83.75 |
| 6 | 83.66666666666667 | // and how can I make this result shorter like 83.67?
| 7 | 80 |
| 9 | 88 |
| 10 | 80.2 |
| 11 | 80 |
| 12 | 99 |
+------------+-------------------+
我试过以下代码
mysql> select student_id, avg(score) as average_score
-> from studentscore
-> group by student_id
-> where avg(score) >= 80;
它给了我一个语法错误。
我知道规则 where 子句应该放在 group by 子句之前,但我不能,因为 where 子句取决于 group by 子句的结果,如果我改变它们的位置,它会给我另一个错误(“群组功能使用无效").
谁能告诉我如何获得我想要的 table ?
最佳答案
用“有”代替“哪里”
关于mysql - 哪里和群冲突了?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43338497/