1.查询每门课的平均成绩：
mysql> select * from course; +----------+-----------+---------+ | cour_num | cour_name | tea_num | +----------+-----------+---------+ | 1-245 | Math | 0438 | | 2-271 | Circuit | 0435 | | 3-105 | OS | 0435 | | 4-321 | Bio | 0436 | +----------+-----------+---------+
mysql> select * from score; +---------+----------+--------+ | stu_num | cour_num | degree | +---------+----------+--------+ | 11422 | 3-105 | 92 | | 11423 | 1-245 | 84 | | 11423 | 2-271 | 75 | | 11424 | 4-321 | 75 | | 11425 | 2-271 | 89 | | 11426 | 1-245 | 61 | | 11426 | 2-271 | 82 | | 11427 | 1-245 | 78 | +---------+----------+--------+
mysql> select avg(degree) from score where cour_num=‘1-245‘; +-------------+ | avg(degree) | +-------------+ | 74.3333 | +-------------+
求平均：avg（）；
mysql> select avg(degree) from score where cour_num=‘2-271‘; +-------------+ | avg(degree) | +-------------+ | 82.0000 | +-------------+ mysql> select avg(degree) from score where cour_num=‘3-105‘; +-------------+ | avg(degree) | +-------------+ | 92.0000 | +-------------+ mysql> select avg(degree) from score where cour_num=‘4-321‘; +-------------+ | avg(degree) | +-------------+ | 75.0000 | +-------------+
在同一条语句中计算4门课程平均值？
mysql> select avg(degree) from score group by cour_num; +-------------+ | avg(degree) | +-------------+ | 74.3333 | | 82.0000 | | 92.0000 | | 75.0000 | +-------------+
mysql> select cour_num,avg(degree) from score group by cour_num; +----------+-------------+ | cour_num | avg(degree) | +----------+-------------+ | 1-245 | 74.3333 | | 2-271 | 82.0000 | | 3-105 | 92.0000 | | 4-321 | 75.0000 | +----------+-------------+
group by：先按课程号分组，分组之后计算平均值。
2.计算score表中至少有2名学生选修的并以2开头的课程的平均分数：
按课程号分组显示：
mysql> select cour_num from score group by cour_num; +----------+ | cour_num | +----------+ | 1-245 | | 2-271 | | 3-105 | | 4-321 | +----------+
至少有2名同学选修：group by 后跟 having
mysql> select cour_num from score group by cour_num -> having count(cour_num)>=2; +----------+ | cour_num | +----------+ | 1-245 | | 2-271 | +----------+
并且要以2开头：like模糊查询
mysql> select cour_num from score group by cour_num -> having count(cour_num)>=2 and cour_num like ‘2%‘; +----------+ | cour_num | +----------+ | 2-271 | +----------+
计算出平均值：avg
mysql> select cour_num,avg(degree) from score group by cour_num -> having count(cour_num)>=2 and cour_num like ‘2%‘; +----------+-------------+ | cour_num | avg(degree) | +----------+-------------+ | 2-271 | 82.0000 | +----------+-------------+
也可知道选修的学生数：count
mysql> select cour_num,avg(degree),count(*) from score group by cour_num -> having count(cour_num)>=2 and cour_num like ‘2%‘; +----------+-------------+----------+ | cour_num | avg(degree) | count(*) | +----------+-------------+----------+ | 2-271 | 82.0000 | 3 | +----------+-------------+----------+
3.查询分数大于70，小于90的stu_num列：
mysql> select stu_num,degree from score where degree>70 and degree<90; +---------+--------+ | stu_num | degree | +---------+--------+ | 11423 | 84 | | 11423 | 75 | | 11424 | 75 | | 11425 | 89 | | 11426 | 82 | | 11427 | 78 | +---------+--------+
mysql> select stu_num,degree from score where degree between 70 and 90; +---------+--------+ | stu_num | degree | +---------+--------+ | 11423 | 84 | | 11423 | 75 | | 11424 | 75 | | 11425 | 89 | | 11426 | 82 | | 11427 | 78 | +---------+--------+
原文：https://www.cnblogs.com/direwolf22/p/11899543.html