学生各门课程成绩统计SQL语句大全(笔试题)

笔试的时候是看一张成绩表，包括姓名，课程，成绩，学号。三个问题，第一，查询某科成绩大于80分的学生。二，学生成绩统计并排名。三，展示一个表格，增加一列，展示每一个学生低于60分就算不及格的，高于60算及格。

具体题目忘记了，从网上找了个相似的练习下。工作上用的比较多的是查询，其它的毕竟好几年没复习了，早就忘了。

成绩表数据如下：

一、查询每个人的总成绩并按从高到低排名(要求显示字段：姓名，总成绩，学号) 

考察聚合函数sum()，用于返回数值列的总数（总额）。。group by 分组（配合聚合函数使用），排序用order by，默认从小到大，加个desc，从大到小。没有where子句查询所有记录

select name,SUM(score),stid from dbo.stuscore
group by name,stid 
order by SUM(score) desc;

运行结果（一块复制进来了）（Navicat命令列界面，输入sql语句按回车enter键）

mysql> select name,SUM(score),stid from dbo.stuscore
group by name,stid 
order by SUM(score) desc;
+------+------------+------+
| name | SUM(score) | stid |
+------+------------+------+
| 张三 | 218        |    1 |
| 李四 | 200        |    2 |
| 王五 | 191        |    3 |
+------+------------+------+
3 rows in set (0.03 sec)

二、查询每个人单科最高成绩(要求显示字段: 学号，姓名，课程，最高成绩)

考察派生表（需要别名），max()函数，用于返回一列中的最大值，NULL 值不包括在计算中。先查学号与最高成绩，再把学号，姓名，课程，最高成绩查出来

select table1.stid,table1.name,table1.subject,table1.score from stuscore table1,
(select stid,max(score) as maxscore from stuscore group by stid) table2 
where table1.stid=table2.stid and table1.score=table2.maxscore;

先查括号里面的

mysql> select stid,max(score) as maxscore from stuscore group by stid;
+------+----------+
| stid | maxscore |
+------+----------+
|    1 |       89 |
|    2 |       80 |
|    3 |       91 |
+------+----------+
3 rows in set (0.03 sec)

结果，再把相关信息列出来

mysql> select table1.stid,table1.name,table1.subject,table1.score from stuscore table1,
(select stid,max(score) as maxscore from stuscore group by stid) table2 
where table1.stid=table2.stid and table1.score=table2.maxscore;
+------+------+---------+-------+
| stid | name | subject | score |
+------+------+---------+-------+
|    1 | 张三 | 数学    |    89 |
|    2 | 李四 | 英语    |    80 |
|    3 | 王五 | 数学    |    91 |
+------+------+---------+-------+
3 rows in set (0.04 sec)

三、查询每个人的平均成绩（要求显示字段: 学号，姓名，平均成绩）

考察avg()函数，用于返回数值列的平均值，NULL 值不包括在计算中。和分组group by用法

select name,stid,avg(score) from dbo.stuscore
group by name,stid;

运行结果

mysql> select name,stid,avg(score) from dbo.stuscore
group by name,stid;
+------+------+------------+
| name | stid | avg(score) |
+------+------+------------+
| 张三 |    1 | 72.6667    |
| 李四 |    2 | 66.6667    |
| 王五 |    3 | 63.6667    |
+------+------+------------+
3 rows in set (0.04 sec)

四、查询各门课程成绩最好的学生(要求显示字段: 学号，姓名，科目，成绩) 

类似第二题。第一步查询把学号 换成科目

select table1.stid,table1.name,table1.subject,table1.score from stuscore table1,
(select subject,max(score) as maxscore from stuscore group by subject) table2 
where table1.subject=table2.subject and table1.score=table2.maxscore;

运行结果

mysql> select table1.stid,table1.name,table1.subject,table1.score from stuscore table1,
(select subject,max(score) as maxscore from stuscore group by subject) table2 
where table1.subject=table2.subject and table1.score=table2.maxscore;
+------+------+---------+-------+
| stid | name | subject | score |
+------+------+---------+-------+
|    2 | 李四 | 语文    |    70 |
|    2 | 李四 | 英语    |    80 |
|    3 | 王五 | 数学    |    91 |
+------+------+---------+-------+
3 rows in set (0.04 sec)

五、查询各门课程成绩最好的两位学生(要求显示字段: 学号，姓名，科目，成绩) 

服了，还没找到更好的方法，将就下。按照最快的思路还没调试好，用top应该更快（找到科目相同的，成绩的，前两列）

SELECT t1.name,t1.SUBJECT,t1.score FROM stuscore t1 WHERE(SELECT COUNT(*) FROM stuscore t2 
WHERE t1.SUBJECT=t2.SUBJECT AND t2.score>=t1.score)<=2 
order by subject desc;

运行结果

mysql> SELECT t1.name,t1.SUBJECT,t1.score FROM stuscore t1 WHERE
(
SELECT COUNT(*) FROM stuscore t2 WHERE
t1.SUBJECT=t2.SUBJECT AND t2.score>=t1.score
)<=2 order by subject desc;
+------+---------+-------+
| name | SUBJECT | score |
+------+---------+-------+
| 李四 | 语文    |    70 |
| 王五 | 语文    |    60 |
| 张三 | 英语    |    70 |
| 李四 | 英语    |    80 |
| 张三 | 数学    |    89 |
| 王五 | 数学    |    91 |
+------+---------+-------+
6 rows in set (0.06 sec)

六、查询结果统计如下

学号	姓名	语文	数学	英语	总分	平均分

考察case when函数。sum(case ...then... else 0 end)固定用法

select stid 学号,name 姓名,sum(case when subject='语文' then score else 0 end )as 语文,
sum(case when subject='数学' then score else 0 end )as 数学,
sum(case when subject='英语' then score else 0 end )as 英语,
SUM(score)总分,avg(score)平均分 from stuscore
group by stid,name order by 总分;

运行结果

mysql> select stid 学号,name 姓名,sum(case when subject='语文' then score else 0 end )as 语文,
sum(case when subject='数学' then score else 0 end )as 数学,
sum(case when subject='英语' then score else 0 end )as 英语,
SUM(score)总分,avg(score)平均分 from stuscore
group by stid,name order by 总分;
+------+------+------+------+------+------+---------+
| 学号 | 姓名 | 语文 | 数学 | 英语 | 总分 | 平均分  |
+------+------+------+------+------+------+---------+
|    3 | 王五 | 60   | 91   | 40   | 191  | 63.6667 |
|    2 | 李四 | 70   | 50   | 80   | 200  | 66.6667 |
|    1 | 张三 | 59   | 89   | 70   | 218  | 72.6667 |
+------+------+------+------+------+------+---------+
3 rows in set (0.05 sec)

七、查询各门课程的平均成绩（要求显示字段：课程，平均成绩）

考察avg函数，group by 分组

select subject,avg(score) from stuscore
group by subject;

运行结果

mysql> select subject,avg(score) from stuscore
group by subject;
+---------+------------+
| subject | avg(score) |
+---------+------------+
| 数学    | 76.6667    |
| 英语    | 63.3333    |
| 语文    | 63.0000    |
+---------+------------+
3 rows in set (0.03 sec)

八、查询数学成绩的排名（要求显示字段：学号，姓名，成绩，排名）

考察count用法，order by排序，注意：比较的次数+1 = 排名

select stid,name,score,
(select count(*) from stuscore t1 where subject ='数学' and t1.score > t2.score)+1 as 名次 from stuscore t2  
where subject='数学' order by score desc;
--注释：排序，比较大小，比较的次数+1 = 排名。

 运行结果

mysql> select stid,name,score,
(select count(*) from stuscore t1 where subject ='数学' and t1.score > t2.score)+1 as 名次 from stuscore t2  
where subject='数学' order by score desc;
+------+------+-------+------+
| stid | name | score | 名次 |
+------+------+-------+------+
|    3 | 王五 |    91 |    1 |
|    1 | 张三 |    89 |    2 |
|    2 | 李四 |    50 |    3 |
+------+------+-------+------+
3 rows in set (0.13 sec)

九、查询数学成绩在2-3名的学生（要求显示字段：学号，姓名，科目，成绩） 

有很多种方法，下面只是其中的一种，顺着第八题做下来

select t3.*  from (
 select stid,name,subject,score,
(select count(*) from stuscore t1 where subject ='数学' and t1.score > t2.score)+1 as 名次 from
 stuscore t2  where subject='数学') t3 
 where t3.名次 between 2 and 3 order by t3.score desc;

运行结果

mysql> select t3.*  from (
 select stid,name,subject,score,
(select count(*) from stuscore t1 where subject ='数学' and t1.score > t2.score)+1 as 名次 from
 stuscore t2  where subject='数学') t3 
 where t3.名次 between 2 and 3 order by t3.score desc;
+------+------+---------+-------+------+
| stid | name | subject | score | 名次 |
+------+------+---------+-------+------+
|    1 | 张三 | 数学    |    89 |    2 |
|    2 | 李四 | 数学    |    50 |    3 |
+------+------+---------+-------+------+
2 rows in set (0.49 sec)

十、查询李四的数学成绩的排名 

方法很多，下面列出其中的一种，顺着第八题做

select stid,name,subject,score,(select count(*) from stuscore t1 where subject ='数学' and t1.score > t2.score)+1 as 名次
 from stuscore t2  where subject='数学' and name = '李四' order by score desc;

运行结果

mysql> select stid,name,subject,score,(select count(*) from stuscore t1 where subject ='数学' and t1.score > t2.score)+1 as 名次
 from stuscore t2  where subject='数学' and name = '李四' order by score desc;
+------+------+---------+-------+------+
| stid | name | subject | score | 名次 |
+------+------+---------+-------+------+
|    2 | 李四 | 数学    |    50 |    3 |
+------+------+---------+-------+------+
1 row in set (0.05 sec)

 

十一、查询统计如下

课程	不及格（0-59）个	良（60-80）个	优（81-100）个

考察case when

 select subject 科目,sum(case when score between 0 and 59 then 1 else 0 end) as 不及格,
 sum(case when score between 60 and 80 then 1 else 0 end) as 良,
 sum(case when score between 81 and 100 then 1 else 0 end) as 优秀 from stuscore
 group by subject;

运行结果

mysql>  select subject 科目,sum(case when score between 0 and 59 then 1 else 0 end) as 不及格,
 sum(case when score between 60 and 80 then 1 else 0 end) as 良,
 sum(case when score between 81 and 100 then 1 else 0 end) as 优秀 from stuscore
 group by subject;
+------+--------+----+------+
| 科目 | 不及格 | 良 | 优秀 |
+------+--------+----+------+
| 数学 | 1      | 0  | 2    |
| 英语 | 1      | 2  | 0    |
| 语文 | 1      | 2  | 0    |
+------+--------+----+------+
3 rows in set (0.06 sec)

十二、查询统计如下

数学: 张三(50分),李四(90分),王五(90分),赵六(76分) 

查了下答案，先贴上

 declare @s nvarchar(1000)
 set @s=''
 select @s =@s+','+name+'('+convert(nvarchar(10),score)+'分)' from 
 stuscore where subject='数学'
 set @s=stuff(@s,1,1,' ')print '数学:'+@s