但行好事
莫论前程❤

Sql语句练习50题(Mysql版)

参考:https://blog.csdn.net/fashion2014/article/details/78826299

SQL各关键字的执行顺序

(8)SELECT(9)DISTINCT <select_list>
(1)FROM <left_table>
(3)<join_type> JOIN <right_table>
(2)ON <join_condition>
(4)WHERE <where_condition>
(5)GROUP BY <grout_by_list>
(6)WITH {CUTE|ROLLUP}
(7)HAVING <having_condition>
(10)ORDER BY <order_by_list>
(11)LIMIT <limit_number>~

表名和字段

  • 学生表

Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别

+------+--------+------------+-------+
| s_id | s_name | s_birth    | s_sex |
+------+--------+------------+-------+
| 01   | 赵雷   | 1990-01-01 | 男    |
| 02   | 钱电   | 1990-12-21 | 男    |
| 03   | 孙风   | 1990-05-20 | 男    |
| 04   | 李云   | 1990-08-06 | 男    |
| 05   | 周梅   | 1991-12-01 | 女    |
| 06   | 吴兰   | 1992-03-01 | 女    |
| 07   | 郑竹   | 1989-07-01 | 女    |
| 08   | 王菊   | 1990-01-20 | 女    |
+------+--------+------------+-------+
8 rows in set (0.00 sec)
  • 课程表

Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号

+------+--------+------+
| c_id | c_name | t_id |
+------+--------+------+
| 01   | 语文   | 02   |
| 02   | 数学   | 01   |
| 03   | 英语   | 03   |
+------+--------+------+
3 rows in set (0.00 sec)
  • 教师表

Teacher(t_id,t_name) –教师编号,教师姓名

+------+--------+
| t_id | t_name |
+------+--------+
| 01   | 张三   |
| 02   | 李四   |
| 03   | 王五   |
+------+--------+
3 rows in set (0.00 sec)
  • 成绩表

Score(s_id,c_id,s_score) –学生编号,课程编号,分数

+------+------+---------+
| s_id | c_id | s_score |
+------+------+---------+
| 01   | 01   |      80 |
| 01   | 02   |      90 |
| 01   | 03   |      99 |
| 02   | 01   |      70 |
| 02   | 02   |      60 |
| 02   | 03   |      80 |
| 03   | 01   |      80 |
| 03   | 02   |      80 |
| 03   | 03   |      80 |
| 04   | 01   |      50 |
| 04   | 02   |      30 |
| 04   | 03   |      20 |
| 05   | 01   |      76 |
| 05   | 02   |      87 |
| 06   | 01   |      31 |
| 06   | 03   |      34 |
| 07   | 02   |      89 |
| 07   | 03   |      98 |
+------+------+---------+
18 rows in set (0.00 sec)

测试数据

--建表
--学生表
CREATE TABLE `Student`(
    `s_id` VARCHAR(20),
    `s_name` VARCHAR(20) NOT NULL DEFAULT '',
    `s_birth` VARCHAR(20) NOT NULL DEFAULT '',
    `s_sex` VARCHAR(10) NOT NULL DEFAULT '',
    PRIMARY KEY(`s_id`)
);
--课程表
CREATE TABLE `Course`(
    `c_id`  VARCHAR(20),
    `c_name` VARCHAR(20) NOT NULL DEFAULT '',
    `t_id` VARCHAR(20) NOT NULL,
    PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
    `t_id` VARCHAR(20),
    `t_name` VARCHAR(20) NOT NULL DEFAULT '',
    PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
    `s_id` VARCHAR(20),
    `c_id`  VARCHAR(20),
    `s_score` INT(3),
    PRIMARY KEY(`s_id`,`c_id`)
);
--插入学生表测试数据
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
--课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');

--教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

--成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);

练习题和sql语句

  • 查询”01″课程比”02″课程成绩高的学生的信息及课程分数(成绩表)

    分析: “01”课程比”02″课程成绩高的 为限定条件

    ​ 先查询学生的信息及课程分数,课程分数中无null值所以使用内连接.限定条件

    ​ student.s_id = score.s_id

    ​ 然后先查出”01″课程的所有学生信息表 再做左连接

select a.* ,b.s_score as 01_score,c.s_score as 02_score from 
Student a join Score b on a.s_id=b.s_id and b.c_id='01'
          left join Score c on a.s_id=c.s_id and c.c_id='02' or c.c_id = NULL where b.s_score>c.s_score;
====================================================================================
+------+--------+------------+-------+----------+----------+
| s_id | s_name | s_birth    | s_sex | 01_score | 02_score |
+------+--------+------------+-------+----------+----------+
| 02   | 钱电   | 1990-12-21 | 男    |       70 |       60 |
| 04   | 李云   | 1990-08-06 | 男    |       50 |       30 |
+------+--------+------------+-------+----------+----------+
2 rows in set (0.00 sec)
  • 也可以这样写
select a.*,b.s_score as 01_score,c.s_score as 02_score from Student a,Score b,Score c 
            where a.s_id=b.s_id 
            and a.s_id=c.s_id 
            and b.c_id='01' 
            and c.c_id='02' 
            and b.s_score>c.s_score
  • 查询”01″课程比”02″课程成绩低的学生的信息及课程分数
select a.* ,b.s_score as 01_score,c.s_score as 02_score from 
    Student a left join Score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL 
     join Score c on a.s_id=c.s_id and c.c_id='02' where b.s_score < c.s_score;
===============================================================================
+------+--------+------------+-------+----------+----------+
| s_id | s_name | s_birth    | s_sex | 01_score | 02_score |
+------+--------+------------+-------+----------+----------+
| 01   | 赵雷   | 1990-01-01 | 男    |       80 |       90 |
| 05   | 周梅   | 1991-12-01 | 女    |       76 |       87 |
+------+--------+------------+-------+----------+----------+
2 rows in set (0.00 sec)
  • 查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩

    分析:平均成绩大于等于60分的

    ​ 同学的学生编号和学生姓名和平均成绩 (avg(s.s_score,2))

    分组聚合 : 聚合要在分组的前提下进行

select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from 
    Student b 
    join Score a on b.s_id = a.s_id
    GROUP BY b.s_id,b.s_name HAVING avg_score >=60;
==============================================================
+------+--------+-----------+
| s_id | s_name | avg_score |
+------+--------+-----------+
| 01   | 赵雷   |     89.67 |
| 02   | 钱电   |     70.00 |
| 03   | 孙风   |     80.00 |
| 05   | 周梅   |     81.50 |
| 07   | 郑竹   |     93.50 |
+------+--------+-----------+
5 rows in set (0.00 sec)
  • 查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩 — (包括有成绩的和无成绩的)

    小于60分包括有成绩和无成绩两种

    无成绩不能使用多表联查.需要单独实现,再做联合查询

select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from 
    Student b 
    left join Score a on b.s_id = a.s_id
    GROUP BY b.s_id,b.s_name HAVING avg_score <60
    union
select a.s_id,a.s_name,0 as avg_score from 
    Student a 
    where a.s_id not in (select distinct s_id from Score);
========================================================================
+------+--------+-----------+
| s_id | s_name | avg_score |
+------+--------+-----------+
| 04   | 李云   |     33.33 |
| 06   | 吴兰   |     32.50 |
| 08   | 王菊   |      0.00 |
+------+--------+-----------+
3 rows in set (0.00 sec)
  • 查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩

    总成绩 选课总数 需要用到聚合函数 –> 分组聚合

    注意: 有些学生没有成绩–>左连接 以学生表为主

select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from 
    Student a 
    left join Score b on a.s_id=b.s_id
    GROUP BY a.s_id,a.s_name;
=====================================================
+------+--------+------------+-----------+
| s_id | s_name | sum_course | sum_score |
+------+--------+------------+-----------+
| 01   | 赵雷   |          3 |       269 |
| 02   | 钱电   |          3 |       210 |
| 03   | 孙风   |          3 |       240 |
| 04   | 李云   |          3 |       100 |
| 05   | 周梅   |          2 |       163 |
| 06   | 吴兰   |          2 |        65 |
| 07   | 郑竹   |          2 |       187 |
| 08   | 王菊   |          0 |      NULL |
+------+--------+------------+-----------+
8 rows in set (0.00 sec)
  • 查询”李”姓老师的数量
select count(t_id) from Teacher where t_name like '李%';
  • 查询学过”张三”老师授课的同学的信息

分析 : 学过”张三”老师授课的 为限定条件

select a.* from 
    Student a 
    join Score b on a.s_id=b.s_id where b.c_id in(
        select c_id from Course where t_id =(
            select t_id from Teacher where t_name = '张三')
    );
==============================================================
+------+--------+------------+-------+
| s_id | s_name | s_birth    | s_sex |
+------+--------+------------+-------+
| 01   | 赵雷   | 1990-01-01 | 男    |
| 02   | 钱电   | 1990-12-21 | 男    |
| 03   | 孙风   | 1990-05-20 | 男    |
| 04   | 李云   | 1990-08-06 | 男    |
| 05   | 周梅   | 1991-12-01 | 女    |
| 07   | 郑竹   | 1989-07-01 | 女    |
+------+--------+------------+-------+
6 rows in set (0.00 sec)
  • 查询没学过”张三”老师授课的同学的信息
select * from 
    Student c 
    where c.s_id not in(
        select a.s_id from Student a join Score b on a.s_id=b.s_id where b.c_id in(
        select a.c_id from Course a join Teacher b on a.t_id = b.t_id where t_name ='张三'));
  • 查询学过编号为”01″并且也学过编号为”02″的课程的同学的信息
select a.* from 
    Student a,Score b,Score c 
    where a.s_id = b.s_id  and a.s_id = c.s_id and b.c_id='01' and c.c_id='02';
===========================================================
+------+--------+------------+-------+
| s_id | s_name | s_birth    | s_sex |
+------+--------+------------+-------+
| 01   | 赵雷   | 1990-01-01 | 男    |
| 02   | 钱电   | 1990-12-21 | 男    |
| 03   | 孙风   | 1990-05-20 | 男    |
| 04   | 李云   | 1990-08-06 | 男    |
| 05   | 周梅   | 1991-12-01 | 女    |
+------+--------+------------+-------+
5 rows in set (0.00 sec)
  • 查询学过编号为”01″但是没有学过编号为”02″的课程的同学的信息

先做学过编号”01″的查询 再在限制条件中去掉”02″的 利用 and 关键字

select a.* from 
    Student a 
    where a.s_id in (select s_id from Score where c_id='01' ) and a.s_id not in(select s_id from Score where c_id='02')
  • 查询没有学全所有课程的同学的信息

分组聚合;利用课程总数 进行筛选

select s.* from Student s 
left join Score s1 on s1.s_id=s.s_id
group by s.s_id having count(s1.c_id)<(select count(*) from Course) 

或者:

select *
from Student
where s_id not in(
    select s_id from Score t1  
    group by s_id having count(*) =(select count(distinct c_id)  from Course));
=====================================================================
+------+--------+------------+-------+
| s_id | s_name | s_birth    | s_sex |
+------+--------+------------+-------+
| 05   | 周梅   | 1991-12-01 | 女    |
| 06   | 吴兰   | 1992-03-01 | 女    |
| 07   | 郑竹   | 1989-07-01 | 女    |
| 08   | 王菊   | 1990-01-20 | 女    |
+------+--------+------------+-------+
4 rows in set (0.00 sec)
  • 查询至少有一门课与学号为”01″的同学所学相同的同学的信息

分析: 学生的信息–> 至少有一门(in) 所学相同课程的学生 –> 与01

​ distinct 去重 因为在score成绩表中 课程与学生的关系是多对多的关系

select * from Student where s_id in(
    select distinct a.s_id from Score a where a.c_id in(select a.c_id from Score a where a.s_id='01')
);
========================================================================
+------+--------+------------+-------+
| s_id | s_name | s_birth    | s_sex |
+------+--------+------------+-------+
| 01   | 赵雷   | 1990-01-01 | 男    |
| 02   | 钱电   | 1990-12-21 | 男    |
| 03   | 孙风   | 1990-05-20 | 男    |
| 04   | 李云   | 1990-08-06 | 男    |
| 05   | 周梅   | 1991-12-01 | 女    |
| 06   | 吴兰   | 1992-03-01 | 女    |
| 07   | 郑竹   | 1989-07-01 | 女    |
+------+--------+------------+-------+
7 rows in set (0.00 sec)
  • 查询和”01″号的同学学习的课程完全相同的其他同学的信息
SELECT
 Student.*
FROM
 Student
WHERE
 s_id IN (SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (
    #下面的语句是找到'01'同学学习的课程数
    SELECT COUNT(c_id) FROM Score WHERE s_id = '01'
   )
 )
AND s_id NOT IN (
 #下面的语句是找到学过‘01’同学没学过的课程,有哪些同学。并排除他们
 SELECT s_id FROM Score
 WHERE c_id IN(
   #下面的语句是找到‘01’同学没学过的课程
   SELECT DISTINCT c_id FROM Score
   WHERE c_id NOT IN (
     #下面的语句是找出‘01’同学学习的课程
     SELECT c_id FROM Score WHERE s_id = '01'
    )
  ) GROUP BY s_id
) #下面的条件是排除01同学
AND s_id NOT IN ('01');
SELECT
 t3.*
FROM
 (
  SELECT
   s_id,
   group_concat(c_id ORDER BY c_id) group1
  FROM
   Score
  WHERE
   s_id &lt;> '01'
  GROUP BY
   s_id
 ) t1
INNER JOIN (
 SELECT
  group_concat(c_id ORDER BY c_id) group2
 FROM
  Score
 WHERE
  s_id = '01'
 GROUP BY
  s_id
) t2 ON t1.group1 = t2.group2
INNER JOIN Student t3 ON t1.s_id = t3.s_id;
  • 查询没学过”张三”老师讲授的任一门课程的学生姓名

没学过任意一门 即 not in (都学过)

select a.s_name from Student a where a.s_id not in (
    select s_id from Score where c_id = 
                (select c_id from course where t_id =(
                    select t_id from Teacher where t_name = '张三')));
  • 查询两门及其以上不及格课程的同学的学号,姓名及其平均成绩
select a.s_id,a.s_name,ROUND(AVG(b.s_score)) from 
    Student a 
    left join Score b on a.s_id = b.s_id
    where a.s_id in(
            select s_id from Score where s_score<60 GROUP BY  s_id having count(1)>=2)
    GROUP BY a.s_id,a.s_name;
  • 检索”01″课程分数小于60,按分数降序排列的学生信息
select a.*,b.c_id,b.s_score from 
    Student a,Score b 
    where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.s_score DESC;
  • 按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
select a.s_id,(select s_score from Score where s_id=a.s_id and c_id='01') as 语文,
                (select s_score from Score where s_id=a.s_id and c_id='02') as 数学,
                (select s_score from Score where s_id=a.s_id and c_id='03') as 英语,
            round(avg(s_score),2) as 平均分 from Score a  GROUP BY a.s_id ORDER BY 平均分 DESC;
SELECT a.s_id,MAX(CASE a.c_id WHEN '01' THEN a.s_score END ) 语文, 
MAX(CASE a.c_id WHEN '02' THEN a.s_score END ) 数学, 
MAX(CASE a.c_id WHEN '03' THEN a.s_score END ) 英语, 
avg(a.s_score),b.s_name FROM Score a JOIN Student b ON a.s_id=b.s_id GROUP BY a.s_id ORDER BY 5 DESC
  • 查询各科成绩最高分、最低分和平均分:以如下形式显示:课程ID,课程name,最高分,最低分,平均分,及格率,中等率,优良率,优秀率 –及格为>=60,中等为:70-80,优良为:80-90,优秀为:>=90
select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
    ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,
    ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,
    ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,
    ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率
    from Score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name
  • 按各科成绩进行排序,并显示排名 — mysql没有rank函数
select a.s_id,a.c_id,
        @i:=@i +1 as i保留排名,
        @k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,
        @score:=a.s_score as Score
    from (
        select s_id,c_id,s_score from Score GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
(select * from (select 
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from Score t2 where t2.s_score>=t1.s_score and t2.c_id='01') rank
FROM Score t1 where t1.c_id='01'
order by t1.s_score desc) t1)
union
(select * from (select 
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from Score t2 where t2.s_score>=t1.s_score and t2.c_id='02') rank
FROM Score t1 where t1.c_id='02'
order by t1.s_score desc) t2)
union
(select * from (select 
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from Score t2 where t2.s_score>=t1.s_score and t2.c_id='03') rank
FROM Score t1 where t1.c_id='03'
order by t1.s_score desc) t3)
  • 查询学生的总成绩并进行排名
select a.s_id,
    @i:=@i+1 as i,
    @k:=(case when @score=a.sum_score then @k else @i end) as rank,
    @score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from Score GROUP BY s_id ORDER BY sum_score DESC)a,
    (select @k:=0,@i:=0,@score:=0)s
  • 查询不同老师所教不同课程平均分从高到低显示
select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from Course a
        left join Score b on a.c_id=b.c_id 
        left join Teacher c on a.t_id=c.t_id
        GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;
  • 查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from Score a,(select @i:=0)s where a.c_id='01'  
                                ORDER BY a.s_score DESC  
            )c
            left join Student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3
            UNION
            select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from Score a,(select @j:=0)s where a.c_id='02'  
                                ORDER BY a.s_score DESC
            )c
            left join Student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3
            UNION
            select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from Score a,(select @k:=0)s where a.c_id='03' 
                                ORDER BY a.s_score DESC
            )c
            left join Student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3;
  • 统计各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比
select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from Score a
                left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
                                            ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比
                                from Score GROUP BY c_id)b on a.c_id=b.c_id
                left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
                                            ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比
                                from Score GROUP BY c_id)c on a.c_id=c.c_id
                left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
                                            ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比
                                from Score GROUP BY c_id)d on a.c_id=d.c_id
                left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
                                            ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
                                from Score GROUP BY c_id)e on a.c_id=e.c_id
                left join Course f on a.c_id = f.c_id
  • 查询学生平均成绩及其名次
select a.s_id,
                @i:=@i+1 as '不保留空缺排名',
                @k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
                @avg_score:=avg_s as '平均分'
        from (select s_id,ROUND(AVG(s_score),2) as avg_s from Score GROUP BY s_id ORDER BY avg_s DESC)a,(select @avg_score:=0,@i:=0,@k:=0)b;
  • 查询各科成绩前三名的记录
    -- 1.选出b表比a表成绩大的所有组
            -- 2.选出比当前id成绩大的 小于三个的
        select a.s_id,a.c_id,a.s_score from Score a 
            left join Score b on a.c_id = b.c_id and a.s_score<b.s_score
            group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3
            ORDER BY a.c_id,a.s_score DESC
  • 查询每门课程被选修的学生数
select c_id,count(s_id) from Score a GROUP BY c_id
  • 查询出只有两门课程的全部学生的学号和姓名
select s_id,s_name from Student where s_id in(
                select s_id from Score GROUP BY s_id HAVING COUNT(c_id)=2);
  • 查询男生、女生人数
select s_sex,COUNT(s_sex) as 人数  from Student GROUP BY s_sex
  • 查询名字中含有”风”字的学生信息
select * from Student where s_name like '%风%';
  • 查询同名同性学生名单,并统计同名人数
select a.s_name,a.s_sex,count(*) from Student a  JOIN 
                    Student b on a.s_id !=b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex
        GROUP BY a.s_name,a.s_sex
  • 查询1990年出生的学生名单
select s_name from Student where s_birth like '1990%'
  • 查询每门课程的平均成绩,结果按平均成绩降序排列,平均成绩相同时,按课程编号升序排列
select c_id,ROUND(AVG(s_score),2) as avg_score from Score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC
  • 查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩
select a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from Score a
        left join Student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85;
  • 查询课程名称为”数学”,且分数低于60的学生姓名和分数
select a.s_name,b.s_score from Score b join Student a on a.s_id=b.s_id where b.c_id=(
                    select c_id from Course where c_name ='数学') and b.s_score<60
  • 查询所有学生的课程及分数情况;
select a.s_id,a.s_name,
                    SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',
                    SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',
                    SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',
                    SUM(b.s_score) as  '总分'
        from Student a left join Score b on a.s_id = b.s_id 
        left join Course c on b.c_id = c.c_id 
        GROUP BY a.s_id,a.s_name
  • 查询任何一门课程成绩在70分以上的姓名、课程名称和分数;
select a.s_name,b.c_name,c.s_score from Course b left join Score c on b.c_id = c.c_id
                left join Student a on a.s_id=c.s_id where c.s_score>=70
  • 查询不及格的课程
select a.s_id,a.c_id,b.c_name,a.s_score from Score a left join Course b on a.c_id = b.c_id
            where a.s_score<60 
  • 查询课程编号为01且课程成绩在80分以上的学生的学号和姓名;
select a.s_id,b.s_name from Score a LEFT JOIN Student b on a.s_id = b.s_id
            where a.c_id = '01' and a.s_score>80
  • 求每门课程的学生人数
select count(*) from Score GROUP BY c_id;
  • 查询选修”张三”老师所授课程的学生中,成绩最高的学生信息及其成绩
-- 查询老师id   
        select c_id from Course c,Teacher d where c.t_id=d.t_id and d.t_name='张三'
        -- 查询最高分(可能有相同分数)
        select MAX(s_score) from Score where c_id='02'
        -- 查询信息
        select a.*,b.s_score,b.c_id,c.c_name from Student a
            LEFT JOIN Score b on a.s_id = b.s_id
            LEFT JOIN Course c on b.c_id=c.c_id
            where b.c_id =(select c_id from Course c,Teacher d where c.t_id=d.t_id and d.t_name='张三')
            and b.s_score in (select MAX(s_score) from Score where c_id='02')
  • 查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
select DISTINCT b.s_id,b.c_id,b.s_score from Score a,Score b where a.c_id != b.c_id and a.s_score = b.s_score
  • 查询每门功成绩最好的前两名
-- 牛逼的写法
    select a.s_id,a.c_id,a.s_score from Score a
        where (select COUNT(1) from Score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id
  • 统计每门课程的学生选修人数(超过5人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,若人数相同,按课程号升序排列
select c_id,count(*) as total from Score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC
  • 检索至少选修两门课程的学生学号
select s_id,count(*) as sel from Score GROUP BY s_id HAVING sel>=2
  • 查询选修了全部课程的学生信息
select * from Student where s_id in(        
            select s_id from Score GROUP BY s_id HAVING count(*)=(select count(*) from Course))
  • 查询各学生的年龄
-- 按照出生日期来算,当前月日 < 出生年月的月日则,年龄减一

    select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') - 
                (case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age from Student;
  • 查询本周过生日的学生
select * from Student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
select * from Student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
  • 查询下周过生日的学生
select * from Student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)
  • 查询本月过生日的学生
select * from Student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)
  • 查询下月过生日的学生
select * from Student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth);
赞(0) 打赏
未经允许不得转载:刘鹏博客 » Sql语句练习50题(Mysql版)
分享到: 更多 (0)

评论 抢沙发

评论前必须登录!

 

觉得文章有用就打赏一下文章作者

支付宝扫一扫打赏

微信扫一扫打赏