sql语句练习题 50题

在网上搜索sql练习题，参考借鉴了https://blog.csdn.net/fashion2014/article/details/78826299

所有的题我都自己做了一遍，敲过了一遍，代码方面应该是没有问题的。

做完之后，对连接，分组，逆向思维，有了更深刻的认识，同时还学到了case语法，关于时间的函数，以及排名的实现

有的语句可能写的比较累赘麻烦，这里是为了练习，就不要在意这些了

 有很多题目，例如算平均数的地方都是有些争议，例如学生压根没选其中某门课，算平均分是按学生自己选的课程，还是按所有的课程数目，这个就看大家怎么去理解。

表结构以及表数据

  -- 学生编号,学生姓名, 出生年月,学生性别
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`)
);
-- 课程编号, 课程名称, 教师编号
CREATE TABLE `Course`(
    `c_id`  VARCHAR(20),
    `c_name` VARCHAR(20) NOT NULL DEFAULT '',
    `t_id` VARCHAR(20) NOT NULL,
    PRIMARY KEY(`c_id`)
);
-- 教师编号,教师姓名
CREATE TABLE `Teacher`(
    `t_id` VARCHAR(20),
    `t_name` VARCHAR(20) NOT NULL DEFAULT '',
    PRIMARY KEY(`t_id`)
);
-- 学生编号,课程编号,分数`course`
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);

-------------------------------------------------------------------------------------------------

-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数

/*(题目说的不清楚，假使这里要查的学生信息只有学生编号，学生姓名,课程分数由于比较的是01和02，只展示01课程和02课程的分数) */

SELECT
t1.`s_id`,
t1.`s_name`,
t2.`s_score` AS 01course,
t3.`s_score` AS 02course
FROM  Student t1 ,Score t2,Score t3
WHERE t1.`s_id`=t2.`s_id`
AND   t1.`s_id`=t3.`s_id`
AND   t2.`c_id`='01'
AND   t3.`c_id`='02'
AND   t2.`s_score`>t3.`s_score`;

/*上面是隐式内连接的写法，把几张表连起来用where筛选，下面是先显式内连接再外连接*/

SELECT
t1.`s_id`,t1.`s_name`,
t2.`s_score` AS 01course,
t3.`s_score` AS 02course
FROM Student t1 
JOIN Score t2 ON t1.`s_id` =t2.`s_id` AND t2.`c_id`='01'
LEFT JOIN Score t3 ON t1.`s_id`=t3.`s_id` AND t3.`c_id`='02' WHERE t2.`s_score`>t3.`s_score` ;

为了方便好理解这个，拆开看下每步求得是什么

 

 

 

 如果不加where的比较大小判断

 

 这样就是有选了课程1的学生，去左连选了课程2的学生。出来的结果就是只选了课程1的学生（例如图中的吴兰）和同时选了01，02课程的学生。原作者在And t3.c_id='02'多加了or  t3.c_id is null,这里我也不清楚为什么要这么写，而且不写的结果也是选了01的和同时选了0102的。

/*第三种写法，采用子查询的方式，大体上也是先分开筛选01，02的，不过感觉写的有点重复，这里是练习，就不在意这些了*/

SELECT  t1.`s_id`,t1.`s_name`, 
t2.`s_score` AS 01course,
t3.s_score AS 02Course
FROM Student t1,(SELECT s1.`s_id` ,s1.`s_score` FROM Score s1 WHERE s1.`c_id`='01' ) t2,
(SELECT s2.`s_id`,s2.`s_score` FROM Score s2 WHERE s2.`c_id`='02') t3
WHERE  t1.`s_id`=t2.s_id AND t1.`s_id`=t3.s_id AND t2.s_score>t3.s_score;

-------------------------------------------------------------------------------------------------

-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数

/*这个跟第一题一样，改变下比较的符号就行了*/

SELECT
t1.`s_id`,
t1.`s_name`,
t2.`s_score` AS 01course,
t3.`s_score` AS 02course
FROM  Student t1 ,Score t2,Score t3
WHERE t1.`s_id`=t2.`s_id`
AND   t1.`s_id`=t3.`s_id`
AND   t2.`c_id`='01'
AND   t3.`c_id`='02'
AND   t2.`s_score`<t3.`s_score`;
-- ------------------
SELECT
t1.`s_id`,t1.`s_name`,
t2.`s_score` AS 01course,
t3.`s_score` AS 02course；
FROM Student t1 
JOIN Score t2 ON t1.`s_id` =t2.`s_id` AND t2.`c_id`='01'
LEFT JOIN Score t3 ON t1.`s_id`=t3.`s_id` AND t3.`c_id`='02' 
WHERE t2.`s_score`<t3.`s_score` ;
-- ------------------
SELECT  t1.`s_id`,t1.`s_name`, 
t2.`s_score` AS 01course,
t3.s_score AS 02Course
FROM Student t1,(SELECT s1.`s_id` ,s1.`s_score` FROM Score s1 WHERE s1.`c_id`='01' ) t2,
(SELECT s2.`s_id`,s2.`s_score` FROM Score s2 WHERE s2.`c_id`='02') t3
WHERE  t1.`s_id`=t2.s_id AND t1.`s_id`=t3.s_id AND t2.s_score<t3.s_score;

-- 3、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

这个题考的是分组，我们先只查学生的id和学生平均成绩大于60的，只在score表上做操作。注意分组后再加筛选条件用having

 SELECT  
 t1.`s_id`,
 AVG(t1.`s_score`) AS avgnumber  
 FROM score t1
 GROUP BY t1.`s_id`
 HAVING AVG(t1.`s_score`)>60;

然后我们再将其和学生表联系起来

 SELECT
 t1.`s_id`,
 t1.`s_name`,
 AVG(t2.`s_score`) AS avgnumber
 FROM Student t1,Score t2 
 WHERE t1.`s_id`=t2.`s_id`
 GROUP BY t2.`s_id`
 HAVING AVG(t2.`s_score`)>60;

结果如下

 

 下面是原作者的，原作者用了保留小数精确度的函数，。然后分组条件那里我觉得写一个就行，毕竟学生id和学生姓名是一一对应关系。

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;

--  5、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

这道题我原本用的内连接，但是用内连接展示的只能是符合条件两表都有的数据。有个数据只在学生表有，但是score表没他的课程和分数记录

SELECT 
t1.`s_id`,t1.`s_name`,
COUNT(t2.`c_id`) AS CourseSum,
SUM(t2.`s_score`) AS ScoreSum
FROM Student t1 ,Score t2 
WHERE t1.`s_id`=t2.`s_id`
GROUP BY t2.`s_id` ;

 

 没有学生王菊的数据，于是采用外连接的方法，除去符合条件的，还包括单表独有的

SELECT 
t1.`s_id`,t1.`s_name`,
IFNULL(COUNT(t2.`c_id`),0)  AS CourseSum,
IFNULL(SUM(t2.`s_score`),0) AS ScoreSum  
FROM student t1 LEFT JOIN score t2
ON t1.`s_id`=t2.`s_id` GROUP BY t2.`s_id`;

 

 -- 6、查询"李"姓老师的数量

SELECT
COUNT(t1.t_id)
FROM (SELECT teacher.`t_id` FROM teacher  WHERE teacher.`t_name` LIKE '李%') t1 ;

SELECT COUNT(t_id) FROM teacher WHERE t_name LIKE '李%';

-- 7、查询学过"张三"老师授课的同学的信息 

/*方式一：通过内连接的方式把4张表连起来，加上筛选条件，比较好理解*/

SELECT 
t1.`s_id`,t1.`s_name`
FROM student t1,course t2,teacher t3,score t4
WHERE t3.`t_name`='张三'
AND t3.`t_id`=t2.`t_id`
AND t2.`c_id`=t4.`c_id`
AND t4.`s_id`=t1.`s_id`;

/*方式二：通过子查询的方式，先根据课程表和老师表选出张三老师教的课程编号，然后在成绩表里去匹配这些编号，进而得到学生信息*/

SELECT 
a.`s_id`,a.`s_name`
FROM  student a JOIN score b 
ON a.`s_id`=b.`s_id`
WHERE b.`c_id` IN
(SELECT  
c.`c_id`
FROM course c JOIN teacher d
ON c.`t_id`=d.`t_id`
WHERE d.`t_name`='张三'
);

/*原作者方式：连续子查询，先求出张三老师的老师编号，根据老师编号找到课程编号，再去score中根据课程编号找到学生的信息*/

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'));

-- 8、查询没学过"张三"老师授课的同学的信息 

/*方式一：由于上题以及知道了学过张三老师的同学信息，那么就把上题作为一个子查询用进来，所有学生的信息，使用not in,如果学生id不在张三课程学生的id里，即为没学过的*/

SELECT 
e.`s_id`,e.`s_name`
FROM student e
WHERE e.`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  
c.`c_id`
FROM course c JOIN teacher d
ON c.`t_id`=d.`t_id`
WHERE d.`t_name`='张三'
)
)

/*方式二：与方式一相同，不过排除方式是使用left join  ，并且连接条件加上右表 为null,这样即是左表独有的数据*/ 

SELECT 
e.`s_id`,e.`s_name`
FROM student e
LEFT JOIN 
(
SELECT 
a.`s_id`
FROM  student a JOIN score b 
ON a.`s_id`=b.`s_id`
WHERE b.`c_id` IN
(SELECT  
c.`c_id`
FROM course c JOIN teacher d
ON c.`t_id`=d.`t_id`
WHERE d.`t_name`='张三'
)
) g ON e.`s_id`=g.`s_id` WHERE g.`s_id` IS NULL ;

/*方式三：连续子查询，根据老师姓名查老师id,根据老师id查课程id，根据课程id查学生id,再排除*/

SELECT 
a.`s_id`,a.`s_name`
FROM student a
WHERE a.`s_id` NOT IN(
  SELECT  b.`s_id`
  FROM student b ,score c WHERE b.`s_id`=c.`s_id` AND c.`c_id` IN(
        SELECT  d.`c_id`
        FROM course d
        WHERE d.`t_id` IN(
             SELECT  teacher.`t_id`
             FROM teacher  
             WHERE   teacher.`t_name`='张三'        
        )
)
)

 --  9 、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

/*方式一：形式上复杂点，但是很好理解，学生表单独和分数表去显式内连接两次，两次的筛选条件分别是课程号为01，02.然后形成的两个结果集再显示内连接下， 连接条件即是同时选了01，02的，因为是同时，所以最外面的s_id和s_name写前面的还是后面的都无所谓*/

SELECT t5.s_id,t5.s_name FROM
(SELECT 
t1.`s_id`, t1.`s_name`
FROM student t1 JOIN  score t2
ON t1.`s_id`=t2.`s_id`
WHERE t2.`c_id` ='01') t6
JOIN (SELECT t3.`s_id`,t3.`s_name`
      FROM student t3 JOIN score t4
      ON t3.`s_id`=t4.`s_id`
      WHERE t4.`c_id`='02') t5 ON t5.`s_id`=t6.`s_id`;

/*方式二：与上面是一个思路，只不过第一次先显式内连接求选了01的学生信息，后面则直接和分数表再内连接，因为同时选了0102的，那么选了01的学生必定在分数表里同时找到02的记录，所以后面直接连分数表，没有像上面连结果集*/

SELECT t3.s_id,t3.s_name 
FROM
(SELECT 
t1.`s_id`, t1.`s_name`
FROM student t1 JOIN  score t2
ON t1.`s_id`=t2.`s_id`
WHERE t2.`c_id` ='01') t3
JOIN score t4 ON t3.s_id=t4.`s_id` AND t4.`c_id`='02'  ;

/*方式三：采用隐式内连接，where 加and把几个连接条件和筛选条件全写上，比较好理解*/

SELECT 
t1.`s_id`,t1.`s_name`
FROM student t1,score t2,score t3
WHERE t1.`s_id`=t2.`s_id` 
AND t2.`c_id`='01' 
AND t1.`s_id`=t3.`s_id` 
AND t3.`c_id`='02' 

-- 10、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

/*方式一：先找出修了01的学生信息，然后找出修了02的学生的信息，not in排除*/

 SELECT     
 t1.`s_id`,t1.`s_name`
 FROM student t1
 JOIN score t2 ON t1.`s_id`=t2.`s_id` 
 WHERE t2.`c_id`='01' AND t1.`s_id` NOT IN
 (
SELECT  t4.`s_id`
FROM student t4
JOIN score t5 ON t4.`s_id`=t5.`s_id`
WHERE t5.`c_id`='02'  
 ) ;

/*方式二：先找出修了01的学生信息，然后直接通过score表找修了02课程的学生信息，因为分数表有学生id，不需要像第一种去做连接*/

 SELECT 
 t1.`s_id`,t1.`s_name`
 FROM  student t1
 WHERE t1.`s_id` 
 IN(
   SELECT  t2.`s_id`
   FROM student t2,score t3
   WHERE t2.`s_id`=t3.`s_id`
   AND t3.`c_id`='01'
   AND t2.`s_id` NOT IN (
      SELECT  t4.`s_id`
      FROM score t4 WHERE t4.`c_id`='02'     
   )
 )

/*方式三：先显式内连接，分别找出修了01，02的，然后两个结果集再左外连接，排除公共的，即令两个集合连接时右边的条件为null,这样就剩下左边独有的*/

SELECT t3.s_id,t3.s_name
FROM 
 (SELECT     
 t1.`s_id`,t1.`s_name`
 FROM student t1
 JOIN score t2 ON t1.`s_id`=t2.`s_id` 
 WHERE t2.`c_id`='01' ) t3
 LEFT JOIN (
SELECT t4.`s_id`,t4.`s_name`
FROM student t4 JOIN score t5
 ON t4.`s_id`=t5.`s_id`
 WHERE t5.`c_id`='02' 
 ) t6 ON t3.s_id=t6.s_id WHERE t6.s_id IS NULL;

/*方式四：原作者提供，查找是否学习0201课程的时候先不用与学生表关联，只需排除，得到学号，最后再从学生表查找学生信息，这里是在练习，所以各种写法都尝试着写写*/

 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')

 -- 11、查询没有学全所有课程的同学的信息 

/*方式一：连接分组，计算课程的数目是否小于课程表的数目，因为包括一门课都没选的，所以要用左连*/

SELECT 
t1.`s_id`,t1.`s_name`,COUNT(t2.c_id)
FROM student t1 LEFT JOIN score t2
ON t1.`s_id`=t2.s_id GROUP BY t1.`s_id`
HAVING COUNT(t2.c_id)<(SELECT COUNT(t3.`c_id`) FROM course t3 )

/*方式二：先单表分组，找出学全了课程的学生id,然后利用学生表去筛选下，和上面是两种思维*/

SELECT 
t1.`s_id`,t1.`s_name`
FROM student t1 WHERE t1.`s_id` NOT IN
(
SELECT    t2.`s_id`
FROM score t2 GROUP BY t2.`s_id` HAVING COUNT(t2.`c_id`)=(SELECT COUNT(t3.c_id)FROM course t3)
)

-- 12、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

/*方式一：先找出01同学学的课程，两表连接，除去01同学本身，*/

SELECT 
DISTINCT  t2.`s_id`,  t2.`s_name`
FROM student t2,score t3
WHERE t2.`s_id`=t3.`s_id`
AND t2.`s_id`!='01' AND t3.`c_id` IN(
SELECT t1.`c_id`
FROM score  t1 WHERE t1.`s_id`='01' ) ;

/*方式二：连续子查询*/

 SELECT t1.`s_id`,t1.`s_name`
 FROM student t1 
 WHERE t1.`s_id`!='01' AND t1.`s_id` IN(
                  SELECT  t2.`s_id` FROM score t2 WHERE t2.`c_id` IN
                   (
                   SELECT t3.`c_id` FROM score t3 WHERE t3.`s_id`='01'
                   ) 
 )  ;

-- 13、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

这个题也是想了很久，一开始想的思路是，在学生表和分数表连接后用in（01学过的课程）然后加个分组判断下数目和01学过课程数相同的，但这种是不对的，例如01学了abc,02学了acd,02有一门课程在01学过的名单里，这就被记录上了，但其实他们的课程不完全相同。

后来想到了在判断课程数的条件下，利用双重否定，即not in(01学过的课程)，这里也不用像in一样纠结是满足一个还是全部之类的问题，只要学生有一门课程不在01的课程id集合里，就一定会被记录的。然后再正常的连接，加条件，学生不在（学过01同学没学过课程的学生集合），而且加上了与01课程数相同的条件，这样求出来的就是与01完全一样的课程的同学

思路分解

-- 01号同学的课程数量 
 SELECT COUNT(t1.`c_id`)FROM  score t1 WHERE t1.`s_id`='01';
-- 01同学所选课程的id
 SELECT  t1.`c_id` FROM score t1 WHERE t1.`s_id`='01';
-- 课程与01同学有不同的同学的id,有一门即会被记录
 SELECT  t2.`s_id`  FROM score t2 WHERE t2.`c_id` NOT IN( SELECT  t1.`c_id` FROM score t1 WHERE t1.`s_id`='01');

解答

SELECT  t3.`s_id`,t3.`s_name`
FROM  student t3 JOIN score t4
ON t3.`s_id`=t4.`s_id` 
WHERE t4.`s_id`!=01
AND t4.`s_id` NOT IN
(SELECT  t2.`s_id`  FROM score t2 WHERE t2.`c_id` NOT IN( SELECT  t1.`c_id` FROM score t1 WHERE t1.`s_id`='01')
) 
GROUP BY t4.`s_id` 
HAVING COUNT(t4.`c_id`)=(SELECT COUNT(t1.`c_id`)FROM  score t1 WHERE t1.`s_id`='01');

/*方式二：来源原作者文章，供大家参考，也是利用的双重否定*/

SELECT
 Student.*
FROM
 Student
WHERE
 s_id IN (SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (
    SELECT COUNT(c_id) FROM Score WHERE s_id = '01'
   )
 )
AND s_id NOT IN (
 SELECT s_id FROM Score
 WHERE c_id IN(
   SELECT DISTINCT c_id FROM Score
   WHERE c_id NOT IN (
     SELECT c_id FROM Score WHERE s_id = '01'
    )
  ) GROUP BY s_id
) 
AND s_id NOT IN ('01')

/*方式三：来源原作者文章，供大家参考*/

SELECT
 t3.*
FROM
 (
  SELECT
   s_id,
   GROUP_CONCAT(c_id ORDER BY c_id) group1
  FROM
   score
  WHERE
   s_id <> '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

-- 14查询没学过张三老师讲授的任何一门课的学生姓名

/*根据课程表和教师表找出张三教的课程的集合，然后通过分数表找出学了张三课程的学生，最后再用学生表not in排除*/

SELECT s1.`s_id`,s1.`s_name`
FROM  student  s1 WHERE s1.`s_id`
NOT IN(
SELECT t4.`s_id`
FROM  score t4
WHERE  t4.`c_id`  IN(
    SELECT t2.`c_id`
    FROM teacher t1,course t2
    WHERE t1.`t_id`=t2.`t_id`
    AND t1.`t_name`='张三' 
))

-- 15、查询两名及其以上不及格课程同学的学号，姓名，及平均成绩

/*方式一：连接分组的同时，把分数限制加上，分组后把符合条件的行筛出来，考的知识点就是count里不能写判断条件，count(*)算行数*/

SELECT    t1.`s_id`,t1.`s_name` ,SUM(t2.`s_score`)/COUNT(t2.`c_id`) AS avgnumber
FROM student t1,score t2
WHERE
t1.`s_id`=t2.`s_id`
AND t2.s_score<60
GROUP BY t2.`s_id`
HAVING  COUNT(*)>=2

/*方式二：先找出两门没及格的学生id，再连接表*/

SELECT t2.`s_id`,t2.`s_name`,SUM(t3.`s_score`)/COUNT(t3.`c_id`) AS avgnumber
FROM student t2,score t3
WHERE t2.`s_id`=t3.`s_id`
AND t3.`s_id` IN(
SELECT  t1.`s_id`
FROM score t1 WHERE t1.`s_score`<60 GROUP BY t1.`s_id` HAVING COUNT(*)>=2
)
GROUP BY t2.s_id

-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息

SELECT t1.`s_id`,t1.`s_name`,t2.`s_score`
FROM student t1,score t2
WHERE t1.`s_id`=t2.`s_id`
AND t2.`s_score`<60
AND t2.`c_id`='01'
ORDER BY t2.`s_score` DESC ;

 -- 17、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

/*如果要带上一门课都没选的学生，连接处改为左连接*/

SELECT  t1.`s_id`,t1.`s_name`,(SELECT  t3.`s_score`  FROM score t3 WHERE t3.s_id=t2.`s_id`  AND t3.`c_id`='01') AS 语文,       
                              (SELECT  t4.s_score    FROM score t4 WHERE t4.s_id=t2.`s_id`  AND t4.c_id  ='02') AS 数学,
                              (SELECT  t5.s_score    FROM score t5 WHERE t5.s_id=t2.s_id    AND t5.c_id  ='03') AS 英语,
                              SUM(t2.s_score)/COUNT(t2.c_id) AS avgnumber
FROM student t1,score t2
WHERE t1.`s_id`=t2.`s_id`       
GROUP BY t1.`s_id`
ORDER BY avgnumber DESC;

-- 18.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

/*这里要学下case语法，简单介绍一下，具体的可以网上搜

case  字段

when 判断条件 then  结果 else  结果

end

*/

SELECT 
t2.`c_id`,
t2.`c_name`,
MAX(t1.`s_score`),
MIN(t1.`s_score`),
AVG(t1.`s_score`),
 100*SUM(CASE WHEN t1.s_score >=60 THEN 1 ELSE 0 END)/COUNT(1) AS 及格率,
 100*SUM(CASE WHEN t1.s_score >=70 AND t1.s_score<80  THEN 1 ELSE 0 END)/COUNT(1) AS 中等率,
 100*SUM(CASE WHEN t1.s_score >=80 AND t1.s_score<90  THEN 1 ELSE 0 END)/COUNT(1) AS 优良率,
 100*SUM(CASE WHEN t1.s_score >=90 THEN 1 ELSE 0 END  )/COUNT(1) AS 优秀率
FROM score t1,course t2    
WHERE t1.`c_id`=t2.`c_id`
GROUP BY t1.`c_id`;

-- 19、按各科成绩进行排序，并显示排名 

/*8.0有rank()函数，8.0一下没有，可以使用用户变量，从这里开始会有好几道关于名次的题目，把这里的三种方式弄清楚就差不多了*/

-- 普通使用

SELECT t1.`s_id`,t1.`c_id`,t1.`s_score`,@r:=@r+1 AS '名次'
FROM score t1 ,(SELECT @r:=0)  q ORDER BY t1.`s_score` DESC ;

 -- 并排名次展示

SELECT t1.`s_id`,t1.`c_id`,t1.`s_score`,(CASE WHEN @temp=t1.`s_score` THEN @r 
                                    WHEN @temp:=t1.`s_score`THEN @r:=@r+1 END) AS '名次'
FROM score t1,(SELECT @r:=0,@temp :=NULL) q ORDER BY t1.`s_score` DESC;

 -- 并排名次跳过

 SELECT  a.s_id,a.c_id,a.s_score,a.名次 FROM
(  SELECT t1.`s_id`,t1.`c_id`,t1.`s_score`,@r:=IF(@temp=t1.`s_score` , @r,@ra) AS '名次' ,@ra:=@ra+1,@temp:=t1.`s_score`FROM score t1,
                                                                            (SELECT @r:=0,@temp:=NULL,@ra:=1)q ORDER BY t1.`s_score`DESC
 ) a            ;

 上面的几种写法都是不分课程，所有的混在一起，因为题目说的也不明确，下面用分开的

  SELECT * FROM
 ( SELECT t1.`s_id`,t1.`c_id`,t1.`s_score`,@r:=@r+1 AS'名次'
  FROM score t1,(SELECT @r:=0)a WHERE t1.`c_id`='01' ORDER BY t1.`s_score`DESC
  ) AS k1
  UNION
  SELECT * FROM
 ( SELECT t2.s_id,t2.`c_id`,t2.s_score,@s:=@s+1 AS '名次'
  FROM score t2, (SELECT @s:=0)b WHERE t2.c_id='02'  ORDER BY t2.s_score DESC
  ) AS k2
  UNION
  SELECT *FROM
  (SELECT  t3.s_id,t3.c_id,t3.s_score,@t:=@t+1 AS '名次'
  FROM score t3,(SELECT @t:=0)c WHERE t3.c_id='03' ORDER BY t3.s_score DESC  
  )AS k3                                             
;

 -- 20、查询学生的总成绩并进行排名

SELECT a.*,@r:=@r+1 AS '名次'
FROM
(SELECT t1.`s_id`,IFNULL(SUM(t1.`s_score`),0) AS sumnumber
FROM score t1
GROUP BY t1.`s_id` 
ORDER BY sumnumber DESC) a,(SELECT @r:=0)b

 -- 21、查询不同老师所教不同课程平均分从高到低显示 

SELECT  t3.`t_name`,t2.`c_name`,AVG(t1.`s_score`) AS avgnumber
FROM score t1,course t2,teacher t3
WHERE t1.`c_id`=t2.`c_id`
AND   t2.`t_id`=t3.`t_id`
GROUP BY t1.`c_id` 
ORDER BY avgnumber DESC;

-- 22、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

/*这个题我直接拿的原作者的，把课程单独分开算的话，其实每个题跟前面几道求名次的题差不多，然后再筛选2-3名，最后三个部份并起来*/

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;

-- 23、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 
t1.`c_id`,t2.`c_name`,
SUM(CASE WHEN t1.`s_score`>85 AND t1.`s_score`<=100 THEN 1 ELSE 0 END)    AS '85-100人数',100*SUM(CASE WHEN t1.`s_score`>85 AND t1.`s_score`<=100 THEN 1 ELSE 0 END) /COUNT(1) AS 占比,
SUM(CASE WHEN t1.s_score  >70 AND t1.s_score  <=85  THEN 1 ELSE 0 END)    AS '70-85人数' ,100*SUM(CASE WHEN t1.s_score  >70 AND t1.s_score  <=85  THEN 1 ELSE 0 END) /COUNT(1) AS 占比,
SUM(CASE WHEN t1.s_score  >60 AND t1.s_score  <=70  THEN 1 ELSE 0 END)    AS '60-70人数' ,100*SUM(CASE WHEN t1.s_score  >60 AND t1.s_score  <=70  THEN 1 ELSE 0 END) /COUNT(1) AS 占比,
SUM(CASE WHEN t1.s_score  >0 AND t1.s_score   <=60  THEN 1 ELSE 0 END)    AS '0-60人数'  ,100*SUM(CASE WHEN t1.s_score  >0 AND t1.s_score   <=60  THEN 1 ELSE 0 END) /COUNT(1) AS 占比
FROM score t1, course t2
WHERE t1.`c_id`=t2.`c_id`
GROUP BY t1.`c_id`;

-- 24、查询学生平均成绩及其名次 

/*分组后再用变量去完成名次功能会出问题，名次排序是跟着id走的，而且变量初始赋值语句不知道写哪合适，所以先分组排序作为一个集合，再进行排序*/

SELECT a.*,@r:=@r+1 AS '名次'
FROM
(SELECT t1.`s_id`,SUM(t1.`s_score`)/COUNT(1) AS avgNumber
FROM score t1
GROUP BY t1.`s_id`
ORDER BY avgNumber DESC )a,(SELECT @r:=0)b

 -- 25、查询各科成绩前三名的记录

/*这个题我直接拿的原作者的答案，还是不太明白*/

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;

-- 26、查询每门课程被选修的学生数

SELECT t1.`c_id`,COUNT(t1.`s_id`) FROM score t1 GROUP BY t1.`c_id`

-- 27、查询出只有两门课程的全部学生的学号和姓名

SELECT t2.`s_id`,t2.`s_name`
FROM score t1,student t2 WHERE
t1.`s_id`=t2.`s_id`
GROUP BY t1.`s_id` 
HAVING COUNT(t1.`c_id`)=2;

-- 28、查询男生、女生人数

SELECT  t1.`s_sex`,COUNT(t1.`s_sex`)
FROM student t1 GROUP BY t1.`s_sex`

-- 29、查询名字中含有"风"字的学生信息

SELECT  t1.`s_id`,t1.`s_name` FROM student t1 WHERE t1.`s_name` LIKE '%风%'

-- 30、查询同名同性别学生名单，并统计同名人数 

/*这个是直接拿的原作者的，题目都不是很明白，同名是同姓还是同姓名，语句倒不是很难理解*/

SELECT t1.`s_name`,t1.`s_sex`,COUNT(1)
FROM student t1,student t2
WHERE t1.`s_id`!=t2.s_id
AND t1.`s_name`=t2.s_name
AND t1.`s_sex`=t2.s_sex
GROUP BY t1.`s_name`,t1.`s_sex`

-- 31、查询1990年出生的学生名单

SELECT t1.`s_id`,t1.`s_name` FROM student t1 WHERE t1.`s_birth` LIKE '1990%';

-- 32、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  t1.`c_id`,AVG(t1.`s_score`) AS avgnumber
FROM score t1 GROUP BY t1.`c_id`
ORDER BY AVG(t1.`s_score`) DESC  ,t1.`c_id` ASC  

-- 33、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  t1.`s_id`,t1.`s_name`,AVG(t2.`s_score`)
FROM student t1,score t2
WHERE t1.`s_id`=t2.`s_id`
GROUP BY t1.`s_id`
HAVING AVG(t2.`s_score`)>=85;

-- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数 

SELECT t1.`s_name`,t2.`s_score` 
FROM student t1,score t2,course t3
WHERE t1.`s_id`=t2.`s_id` 
AND t2.`c_id`=t3.c_id
AND t3.c_name='数学'
AND t2.`s_score`<60;

-- 35、查询所有学生的课程及分数情况； 

/*这个我是按着原作者的思路写的，每门课的成绩展示，用sum和case组合起来还是很巧妙的*/

SELECT  t1.`s_name`,
SUM(CASE  WHEN t3.`c_name`='语文' THEN t2.`s_score` ELSE 0 END  ) AS '语文',
SUM(CASE  WHEN t3.`c_name`='数学' THEN t2.`s_score` ELSE 0 END  ) AS '数学',
SUM(CASE  WHEN t3.`c_name`='英语' THEN t2.`s_score` ELSE 0 END  ) AS '英语'                 
FROM student t1 LEFT JOIN score t2 
ON t1.`s_id`=t2.`s_id`
LEFT JOIN course t3 
ON t2.`c_id`=t3.`c_id`
GROUP BY t1.`s_id`

-- 36、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 t2.`s_name`,t3.c_name,t1.`s_score`
FROM score t1,student t2,course t3
WHERE t1.`s_id`=t2.`s_id`
AND t1.`c_id`=t3.c_id
AND  t1.`s_score`>70;

-- 37、查询不及格的课程

SELECT t2.`s_name`,t3.c_name,t1.`s_score`
FROM score t1,student t2,course t3
WHERE t1.`s_id`=t2.`s_id`
AND t1.`c_id`=t3.c_id
AND  t1.`s_score`<60;

-- 38、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 t2.`s_id`, t2.`s_name`
FROM score t1,student t2
WHERE t1.`s_id`=t2.`s_id`
AND  t1.`c_id`='01'
AND t1.`s_score`>=80;

-- 39、求每门课程的学生人数 

SELECT t1.`c_id`,COUNT(*) AS 选课人数
FROM score t1
GROUP BY t1.`c_id`;

-- 40、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

/*张三老师教的课程id*/

SELECT  t1.`c_id` FROM course t1,teacher t2 WHERE t1.`t_id`=t2.`t_id` AND t2.`t_name`='张三';

SELECT  t3.s_name,t4.`s_score`
FROM student t3,score t4
WHERE t3.s_id=t4.`s_id` AND t4.`c_id` IN (SELECT  t1.`c_id` FROM course t1,teacher t2 WHERE t1.`t_id`=t2.`t_id` AND t2.`t_name`='张三')
ORDER BY t4.`s_score` DESC LIMIT 0,1;

-- 41、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 t1.`s_id`,t1.`c_id`,t1.`s_score`
FROM score t1,score t2
WHERE t1.`c_id`!=t2.`c_id`
AND   t1.`s_score`=t2.`s_score`;

-- 42、查询每门功成绩最好的前两名 

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;

-- 43、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  t1.c_id,COUNT(*) AS learnNumber FROM score t1 GROUP BY t1.c_id HAVING learnNumber>5 ORDER BY learnNumber DESC,c_id ASC;

-- 44、检索至少选修两门课程的学生学号 

SELECT  t2.`s_id` ,COUNT(1) FROM score t2  GROUP BY t2.`s_id` HAVING COUNT(1)>=2 ;

-- 45、查询选修了全部课程的学生信息 

 SELECT t1.`s_id`,t1.`s_name`
   FROM student t1,score t2
   WHERE t1.`s_id`=t2.`s_id`
   GROUP BY t1.`s_id` HAVING COUNT(1)= ( SELECT COUNT(1) FROM course  )

-- 46、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)

/*初次接触关于时间的题目，可以跳过这个题，看看后面的题，摸清关键字怎么用再来看这个题，同时这个题借助case语法*/

SELECT t1.`s_name`, t1.`s_birth`,
                          (DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(t1.`s_birth`,'%Y')
                         -( CASE WHEN DATE_FORMAT(NOW(),'%m%d')<DATE_FORMAT(t1.`s_birth`,'%m%d')  
                            THEN 1 ELSE 0 END) )AS age
    FROM student t1

-- 47、查询本周过生日的学生

/*weak函数，计算日期是本年的第几个星期*/

SELECT WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

SELECT t1.`s_id`,t1.`s_name`,t1.`s_birth`
FROM student t1
WHERE WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(t1.`s_birth`);

-- 48、查询下周过生日的学生

SELECT t1.`s_id`,t1.`s_name`,t1.`s_birth` 
FROM student t1
WHERE WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1=WEEK(t1.`s_birth`)

-- 49、查询本月过生日的学生

SELECT t1.`s_id`,t1.`s_name`,t1.`s_birth`
FROM student t1
WHERE MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))=MONTH(t1.`s_birth`);

-- 50、查询下月过生日的学生

SELECT  t1.`s_id`,t1.`s_name`,t1.`s_birth` 
FROM student t1
WHERE MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1=MONTH(t1.`s_birth`);