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`)
);
CREATE TABLE `Course`(
`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);
CREATE TABLE `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
);
CREATE TABLE `Score`(
`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);
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');
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
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);
#题目1 查询课程编号为“01”的课程比“02”的课程成绩高的所有学生的学号（重点）
SELECT a.s_id 's_no',c.s_name,a.s_score '01',b.s_score '02' FROM
(
SELECT s_id,c_id,s_score FROM score WHERE c_id='01'
) as a
INNER JOIN
(
SELECT s_id,c_id,s_score FROM score WHERE c_id='02'
) as b ON a.s_id = b.s_id
INNER JOIN student AS c ON c.s_id = a.s_id
WHERE a.s_score > b.s_score
#INNER JOIN 是交集
#LEFT JOIN 是左连接
#2、查询平均成绩大于60分的学生的学号和平均成绩（简单，第二道重点）
SELECT s_id,AVG(s_score) '平均成绩'
FROM score
GROUP BY s_id HAVING AVG(s_score)>60
#3、查询所有学生的学号、姓名、选课数、总成绩（不重要）
SELECT a.s_id,a.s_name,COUNT(b.c_id),
SUM(CASE
WHEN b.s_score is NULL THEN 0 ELSE b.s_score END)
FROM student as a
LEFT JOIN score as b ON a.s_id = b.s_id
GROUP BY s_id,a.s_name
#4.查询姓“猴”的老师的个数（不重要）
SELECT COUNT(t_id)
FROM teacher
-- WHERE t_name LIKE'猴%'
WHERE t_name LIKE'张%'#%表示任意字符
SELECT COUNT(DISTINCT t_name)#DISTINCT去重的功能
FROM teacher
-- WHERE t_name LIKE'猴%'
WHERE t_name LIKE'张%'#%表示任意字符
#5.查询没学过“张三”老师课的学生的学号、姓名（重点）
#方法1
SELECT s_id,s_name FROM student
WHERE 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'
)
)
)
#方法2
SELECT s_id,s_name FROM student
WHERE s_id NOT IN
(SELECT s_id FROM score as s
INNER JOIN course as c ON s.c_id=c.c_id
INNER JOIN teacher as t ON c.t_id=t.t_id
WHERE t.t_name='张三')
#6.查询学过“张三”老师所教的所有课的同学的学号、姓名（重点）
#方法1
SELECT s_id,s_name FROM student
WHERE s_id 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'
)))
#方法2
SELECT st.s_id,st.s_name,t_name FROM student as st
INNER JOIN score AS s ON s.s_id=st.s_id
INNER JOIN course AS c ON s.c_id=c.c_id
INNER JOIN teacher as t ON t.t_id=c.t_id
WHERE t.t_name= "张三"
ORDER BY st.s_id
#7.查询学过编号为“01”的课程并且也学过编号为“02”的课程的学生的学号、姓名（重点）
SELECT s_id,s_name FROM student
WHERE s_id IN
(SELECT a.s_id FROM #只说s_id会造成模糊不清，不知道是a还是b的
(SELECT s_id FROM score WHERE c_id = '01') AS a
INNER JOIN
(SELECT s_id FROM score WHERE c_id = '02') AS b
ON a.s_id=b.s_id)
-- SELECT LENGTH(NULL)
-- SELECT LENGTH('12345')
#8.查询课程编号为“02”的总成绩（不重点）
# 看谁 就 SELECT 谁！！！
SELECT SUM(s_score),AVG(s_score),COUNT(s_id),COUNT(DISTINCT s_id) FROM score WHERE c_id='02'
# 按照c_id分组求值
SELECT c_id,SUM(s_score),AVG(s_score),COUNT(s_id),COUNT(DISTINCT s_id) FROM score GROUP BY c_id
#HAVING作进一步筛选
SELECT c_id,SUM(s_score),AVG(s_score),COUNT(s_id),COUNT(DISTINCT s_id) FROM score GROUP BY c_id HAVING c_id='02'
#9、查询所有课程成绩小于60分的学生的学号、姓名
SELECT a.s_id,t.s_name FROM
(SELECT s_id,COUNT(DISTINCT c_id) 'cnt'
FROM score WHERE s_score<60 GROUP BY s_id) AS a
INNER JOIN
(SELECT s_id,COUNT(DISTINCT c_id) 'cnt'
FROM score GROUP BY s_id) AS b
ON a.s_id = b.s_id
INNER JOIN
student AS t
ON a.s_id=t.s_id
WHERE a.cnt=b.cnt
#等同于 SELECT s_id,COUNT(DISTINCT c_id) AS cnt
-- 10.查询没有学全所有课的学生的学号、姓名(重点)
#没有考虑一门课都没选的学生
#方法1
SELECT b.s_id,b.s_name FROM
(SELECT s_id,COUNT(DISTINCT c_id) FROM score GROUP BY s_id HAVING COUNT(DISTINCT c_id)<(SELECT COUNT(DISTINCT c_id) FROM course))
AS a
INNER JOIN student
AS b
ON a.s_id = b.s_id
#方法2
SELECT s_id,s_name FROM student
WHERE s_id IN
(SELECT s_id FROM score GROUP BY s_id HAVING (COUNT(DISTINCT c_id)<(SELECT COUNT(DISTINCT c_id) FROM course)))
#考虑了一门课都没选的学生
SELECT * FROM
student AS st
LEFT JOIN
score AS a
ON st.s_id = a.s_id GROUP BY st.s_id HAVING COUNT(DISTINCT c_id)<(SELECT COUNT(DISTINCT c_id) FROM course)
#拓展 看a里面哪些数据是b里面没有的，然后将数据插入到b里面去。
-- INSERT INTO b
-- SELECT * FROM a LEFT JOIN b ON a.id=b.id
-- WHERE b.id is NULL
-- 11、查询至少有一门课与学号为“01”的学生所学课程相同的学生的学号和姓名（重点）
#方法1
SELECT b.s_id,d.c_name,b.s_name FROM
(SELECT s.s_id,s.c_id,a.s_name FROM (
SELECT s_id,c_id FROM score WHERE c_id IN (SELECT c_id FROM score WHERE s_id='01') GROUP BY s_id )
AS s
LEFT JOIN student AS a ON s.s_id = a.s_id ) AS b
LEFT JOIN course AS d ON d.c_id = b.c_id
#方法2
SELECT s_id,s_name FROM student
WHERE s_id IN
(SELECT DISTINCT s_id FROM score WHERE c_id IN
(SELECT c_id FROM score WHERE s_id='01') AND s_id !='01')
#方法3
SELECT a.s_id,a.s_name FROM student as a
INNER JOIN
(SELECT DISTINCT s_id FROM score WHERE c_id IN
(SELECT c_id FROM score WHERE s_id='01') AND s_id !='01')
AS b ON a.s_id=b.s_id
-- 12.查询和“01”号同学所学课程完全相同的其他同学的学号(重点)
SELECT * FROM student
WHERE s_id IN (
SELECT s_id FROM score WHERE s_id !='01' GROUP BY s_id HAVING (COUNT(DISTINCT c_id)=(SELECT COUNT(DISTINCT c_id) FROM score WHERE s_id='01')))
AND s_id NOT IN (
SELECT s_id FROM score WHERE c_id NOT IN (SELECT c_id FROM score WHERE s_id='01')
)
-- 15、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（重点）
SELECT a.s_id,b.s_name,AVG(s_score) FROM (score AS a INNER JOIN student AS b ON a.s_id=b.s_id)
WHERE a.s_score < 60 GROUP BY a.s_id HAVING COUNT(DISTINCT a.c_id)>=2
-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息（和34题重复，不重点）
SELECT * FROM (score AS a INNER JOIN student AS b ON a.s_id=b.s_id)
WHERE (c_id='01' AND s_score <60) ORDER BY a.s_score DESC
#升序排列ASC
-- 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(重重点与35一样)
SELECT
s_id '学号',
MAX(CASE WHEN c_id='01' THEN s_score ELSE NULL END) '语文',
MAX(CASE WHEN c_id='02' THEN s_score ELSE NULL END) '数学',
MAX(CASE WHEN c_id='03' THEN s_score ELSE NULL END) '英语',
AVG(s_score) '平均成绩'
FROM score GROUP BY s_id ORDER BY AVG(s_score) 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 (超级重点)
SELECT s.c_id,
c.c_name,
MAX(s_score),MIN(s_score),AVG(s_score),
SUM(CASE WHEN s.s_score>=60 THEN 1 ELSE 0 END)/COUNT(s_id) '及格',
SUM(CASE WHEN s.s_score>=70 AND s.s_score <80 THEN 1 ELSE 0 END)/COUNT(s_id) '中等',
SUM(CASE WHEN s.s_score>=80 AND s.s_score <90 THEN 1 ELSE 0 END)/COUNT(s_id) '优良',
SUM(CASE WHEN s.s_score>=90 THEN 1 ELSE 0 END)/COUNT(s_id) '优秀'
FROM score AS s INNER JOIN course AS c ON s.c_id=c.c_id GROUP BY c_id
-- 19、按各科成绩进行排序，并显示排名(重点row_number)
-- row_number(）over （order by 列）
# row_number() 即便有相同的值，也会按顺寻排列标示
# dense_rank() 会有 1 2 2 3的顺序
# RANK() 会有 1 2 2 4的顺序
SELECT * ,RANK() OVER (ORDER BY s_score DESC) ranking FROM score
SELECT * ,RANK() OVER (PARTITION by c_id ORDER BY s_score DESC) ranking FROM score
SELECT * ,RANK() OVER (PARTITION by c_id ORDER BY s_score DESC) AS 'rank' FROM score
-- 20、查询学生的总成绩并进行排名（不重点）
#方法1
SELECT s_id,SUM(s_score) FROM score GROUP BY s_id ORDER BY SUM(s_score) DESC
#方法2 #命名的中文字段加不加''都一样
SELECT s_id,SUM(s_score) 总成绩 FROM score GROUP BY s_id ORDER BY 总成绩 DESC
-- 21 、查询不同老师所教不同课程平均分从高到低显示(不重点)
-- 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（重要 25类似）
-- 23、使用分段[100-85],[85-70],[70-60],[<60]来统计各科成绩，分别统计各分数段人数：课程ID和课程名称(重点和18题类似)
SELECT
a.c_id,a.c_name,
SUM(case WHEN 85<s_score AND s_score<=100 THEN 1 ELSE 0 END) 优秀,
SUM(case WHEN 70<s_score AND s_score<=85 THEN 1 ELSE 0 END) 良好,
SUM(case WHEN 60<= s_score AND s_score<=70 THEN 1 ELSE 0 END) 及格,
SUM(case WHEN s_score <60 THEN 1 ELSE 0 END) 不及格
FROM (course AS a LEFT JOIN score AS b ON a.c_id=b.c_id) GROUP BY a.c_id
-- 24、查询学生平均成绩及其名次（同19题，重点）
#错误写法 PARTITION by s_id 表示将s_id化为了一组。
SELECT s_id,AVG(s_score),RANK() OVER ( ORDER BY AVG(s_score) DESC) AS 'rank' FROM score GROUP BY s_id
#上下区别 下边一个将各科分数和人都拆分化，可以进行拆分表示，而上边由于求各人的平均分，所以不能再 PARTITION by 分类排序了，不然分组都是排名为1！！！
SELECT * ,RANK() OVER (PARTITION by c_id ORDER BY s_score DESC) AS 'rank' FROM score
-- 25、查询各科成绩前三名的记录（不考虑成绩并列情况）（重点 与22题类似）我是 “是努力做好产品战略的小王同学呀” 收场、关注，技术、职场不迷路~
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