MySQL数据库练习-4

查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和

分析:这道题明显也只用到sc和student两张表,对这两张表进行联合查询,选课总数和课程成绩总和的操作对象都是sc表,因此我们可以构建表r,里面字段是选课总数和成绩总和还有SId,然后跟student表进行联合查询

构建表r

select sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber from sc 
group by sc.sid;

查询结果:

+------+----------+--------------+
| sid  | scoresum | coursenumber |
+------+----------+--------------+
| 01   |    269.0 |            3 |
| 02   |    210.0 |            3 |
| 03   |    240.0 |            3 |
| 04   |    100.0 |            3 |
| 05   |    163.0 |            2 |
| 06   |     65.0 |            2 |
| 07   |    187.0 |            2 |
+------+----------+--------------+

与student进行联合查询:

select student.sid, student.sname,r.coursenumber,r.scoresum
from student,
(select sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber from sc 
group by sc.sid)r
where student.sid = r.sid;

查询结果:

+------+--------+--------------+----------+
| sid  | sname  | coursenumber | scoresum |
+------+--------+--------------+----------+
| 01   | 赵雷   |            3 |    269.0 |
| 02   | 钱电   |            3 |    210.0 |
| 03   | 孙风   |            3 |    240.0 |
| 04   | 李云   |            3 |    100.0 |
| 05   | 周梅   |            2 |    163.0 |
| 06   | 吴兰   |            2 |     65.0 |
| 07   | 郑竹   |            2 |    187.0 |
+------+--------+--------------+----------+

到这里结果其实就出来了,但是,这里有一点需要思考的地方,我们的同学其实有13个,但是这里只出现了7个,说明有一些同学是没选课的,这里的查询显然是忽略了这一点,所以如果严谨一些,我们需要吧其他没选课的同学都考虑进去,我们用join就可以给没选课的同学一个null值.

先查询student表拿出所有的学生的sid和sname

select student.sid,student.sname from student;

查询结果;

+------+--------+
| sid  | sname  |
+------+--------+
| 01   | 赵雷   |
| 02   | 钱电   |
| 03   | 孙风   |
| 04   | 李云   |
| 05   | 周梅   |
| 06   | 吴兰   |
| 07   | 郑竹   |
| 09   | 张三   |
| 10   | 李四   |
| 11   | 李四   |
| 12   | 赵六   |
| 13   | 孙七   |
+------+--------+

然后拿出入第一种方法一样查询出总分和科目数

select sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber from sc 
       group by sc.sid;

查询结果:

+------+----------+--------------+
| sid  | scoresum | coursenumber |
+------+----------+--------------+
| 01   |    269.0 |            3 |
| 02   |    210.0 |            3 |
| 03   |    240.0 |            3 |
| 04   |    100.0 |            3 |
| 05   |    163.0 |            2 |
| 06   |     65.0 |            2 |
| 07   |    187.0 |            2 |
+------+----------+--------------+

然后left join这两个查询结果,进行查询:

select s.sid, s.sname,r.coursenumber,r.scoresum
from (
    (select student.sid,student.sname 
    from student
    )s 
    left join 
    (select 
        sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber
        from sc 
        group by sc.sid
    )r 
   on s.sid = r.sid
);

最终结果:

+------+--------+--------------+----------+
| sid  | sname  | coursenumber | scoresum |
+------+--------+--------------+----------+
| 01   | 赵雷   |            3 |    269.0 |
| 02   | 钱电   |            3 |    210.0 |
| 03   | 孙风   |            3 |    240.0 |
| 04   | 李云   |            3 |    100.0 |
| 05   | 周梅   |            2 |    163.0 |
| 06   | 吴兰   |            2 |     65.0 |
| 07   | 郑竹   |            2 |    187.0 |
| 09   | 张三   |         NULL |     NULL |
| 10   | 李四   |         NULL |     NULL |
| 11   | 李四   |         NULL |     NULL |
| 12   | 赵六   |         NULL |     NULL |
| 13   | 孙七   |         NULL |     NULL |
+------+--------+--------------+----------+

以上

猜你喜欢