categories

Mysql50句练习上

前言: 故不积跬步，无以至千里

安装地址

https://www.runoob.com/mysql/mysql-install.html

scrpit:

#!/bin/sh

wget http://repo.mysql.com/mysql-community-release-el7-5.noarch.rpm

rpm -ivh mysql-community-release-el7-5.noarch.rpm

sudo yum update -y
sudo yum install mysql-server -y

sudo chown mysql:mysql -R /var/lib/mysql

mysqld --initialize

systemctl start mysqld

systemctl status mysqld

sql 答案参考这里

 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数	
 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
 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
 4、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
 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
 6、查询"李"姓老师的数量 
 7、查询学过"张三"老师授课的同学的信息 
 8、查询没学过"张三"老师授课的同学的信息 
 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
 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
 11、查询没有学全所有课程的同学的信息 
 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 
 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 
 14、查询没学过"张三"老师讲授的任一门课程的学生姓名 
 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 
 16、检索"01"课程分数小于60，按分数降序排列的学生信息
 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
 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
 19、按各科成绩进行排序，并显示排名
 20、查询学生的总成绩并进行排名
 21、查询不同老师所教不同课程平均分从高到低显示 
 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
 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
 24、查询学生平均成绩及其名次 
 25、查询各科成绩前三名的记录

2019.7.23
– 1、查询”01”课程比”02”课程成绩高的学生的信息及课程分数

1. 个人思路：
    先得出01课比02课成绩高的s_id, 再join student表
    SELECT * FROM student st JOIN
    (SELECT
	   a.s_id ,a.s_score as 01sorce, b.s_score as 02sorce
    FROM
	score a
	JOIN score b ON 
	a.s_id = b.s_id
	AND a.c_id = '01'
	AND b.c_id = '02'
	AND a.s_score > b.s_score GROUP BY a.s_id) sc 
	ON st.s_id = sc.s_id;

缺点：使用了子查询，耗时大， 使用了0.002s
答案思路：
    使用三个对象进行比较
    SELECT
	st.*, a.s_score as 01_score, b.s_score as 02_score
    FROM
	student st,
	score a,
	score b 
    WHERE
	a.s_id = st.s_id
	AND b.s_id = st.s_id
	AND a.c_id = '01'
	AND b.c_id = '02'
	AND a.s_score > b.s_score
    信息中只用了0.001s
学到的要点：1.join on的一个使用
            2.条件后必须要是要用到and连接
            3.可以from 三个对象
            4.不要用子查询，会慢一倍

– 2、查询”01”课程比”02”课程成绩低的学生的信息及课程分数

SELECT
	st.*,
	sc.c1,
	sc.c2 
FROM
	Student st
	RIGHT JOIN (
	SELECT
		a.s_id,
		a.s_score AS c1,
		b.s_score AS c2 
	FROM
		( SELECT * FROM Score WHERE c_id = '01' ) AS a
		JOIN ( SELECT * FROM Score WHERE c_id = '02' ) AS b ON a.s_id = b.s_id 
		AND a.s_score < b.s_score 
	) sc ON st.s_id = sc.s_id

– 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

SELECT
	st.*,
	avg.avg 
FROM
	Student st
	INNER JOIN ( SELECT s_id, avg( s_score ) AS avg FROM Score GROUP BY s_id ) avg ON st.s_id = avg.s_id 
	AND avg.avg >= 60 
ORDER BY
	avg.avg DESC
``` 
--  4、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

SELECT
st.*,
avg.avg
FROM
Student st
INNER JOIN ( SELECT s_id, avg( s_score ) AS avg FROM Score GROUP BY s_id ) avg ON st.s_id = avg.s_id
AND avg.avg < 60
ORDER BY
avg.avg DESC

1	--- 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

SELECT
s_name,
res.*
FROM
Student st
LEFT JOIN ( SELECT s_id, COUNT( c_id ) AS ‘选课总数’, SUM( s_score ) AS ‘课程总成绩’ FROM Score GROUP BY s_id ) res ON st.s_id = res.s_id

1 2	-- 20200422 (6-10) -- 6、查询"李"姓老师的数量

SELECT
COUNT( DISTINCT t_id )
FROM
Teacher
WHERE
t_name LIKE ‘李%’

1	-- 7、查询学过"张三"老师授课的同学的信息

SELECT
st.*
FROM
Student st,
Score sc,
Teacher tc,
Course cs
WHERE
tc.t_id = cs.t_tid
AND tc.t_name = ‘张三’
AND sc.c_id = cs.c_id
AND st.s_id = sc.s_id

1	-- 8、查询没学过"张三"老师授课的同学的信息

SELECT
st.*
FROM
Student st
WHERE
st.s_id NOT IN (
SELECT
sc.s_id
FROM
Score sc,
Teacher tc,
Course cs
WHERE
tc.t_id = cs.t_tid
AND tc.t_name = ‘张三’
AND sc.c_id = cs.c_id)

1	-- 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

SELECT
st.*
FROM
Score a,
Score b,
Student st
WHERE
a.c_id = ‘01’
AND b.c_id = ‘02’
AND a.s_id = b.s_id
AND st.s_id = a.s_id
AND st.s_id = b.s_id

1	-- 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

SELECT
st.*
FROM
Student st,
Score sc
WHERE
sc.s_id NOT IN ( SELECT s_id FROM Score WHERE c_id = ‘02’ )
AND sc.s_id = st.s_id
GROUP BY
sc.s_id

1	-- 11、查询没有学全所有课程的同学的信息

SELECT
st.*
FROM
Student st
LEFT JOIN Score sc ON st.s_id = sc.s_id
GROUP BY
sc.s_id
HAVING
COUNT( sc.c_id ) <3

1	-- 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

SELECT
a.s_id
FROM
Student st,
Score a,
Score b
WHERE
a.s_id = st.s_id
AND st.s_id <> ‘01’
AND b.s_id = ‘01’
AND a.c_id = b.c_id
GROUP BY
a.s_id
HAVING
COUNT( a.c_id ) = 3

1	-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT
st1.*
FROM
Student st1
WHERE
st1.s_id NOT IN (
SELECT
st.s_id
FROM
Student st,
Score sc,
Course cs,
Teacher tc
WHERE
st.s_id = sc.s_id
AND sc.c_id = cs.c_id
AND tc.t_id = cs.t_tid
AND tc.t_name = ‘张三’
);

1	-- 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

SELECT
st.*,
avg( sc.s_score ) AS avg,
COUNT( sc.c_id ) AS count
FROM
Score sc,
Student st
WHERE
sc.s_id = st.s_id
AND s_score < 60 GROUP BY s_id HAVING count >=2

1	-- 21、查询不同老师所教不同课程平均分从高到低显示

SELECT
c_id,
AVG( s_score ) AS avg
FROM
Teacher tc,
Score
GROUP BY
c_id
ORDER BY
avg DESC

1	-- 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

SELECT
*
FROM
(
SELECT
st.s_id,
st.s_name,
st.s_sex,
st.s_birth,
sc.c_id,
sc.s_score
FROM
Student st,
Score sc
WHERE
st.s_id = sc.s_id
AND sc.c_id = ‘01’
ORDER BY
sc.s_score DESC
LIMIT 1,
2
) a UNION ALL
SELECT
*
FROM
(
SELECT
st.s_id,
st.s_name,
st.s_sex,
st.s_birth,
sc.c_id,
sc.s_score
FROM
Student st,
Score sc
WHERE
st.s_id = sc.s_id
AND sc.c_id = ‘02’
ORDER BY
sc.s_score DESC
LIMIT 1,
2
) b UNION ALL
SELECT
*
FROM
(
SELECT
st.s_id,
st.s_name,
st.s_sex,
st.s_birth,
sc.c_id,
sc.s_score
FROM
Student st,
Score sc
WHERE
st.s_id = sc.s_id
AND sc.c_id = ‘03’
ORDER BY
sc.s_score DESC
LIMIT 1,
2
) c

1	-- 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
c.c_id,
c.c_name,((
SELECT
count( 1 )
FROM
Score sc
WHERE
sc.c_id = c.c_id
AND sc.s_score <= 100 AND sc.s_score > 80
)/(
SELECT
count( 1 )
FROM
Score sc
WHERE
sc.c_id = c.c_id
)) “100-85”,((
SELECT
count( 1 )
FROM
Score sc
WHERE
sc.c_id = c.c_id
AND sc.s_score <= 85 AND sc.s_score > 70
)/(
SELECT
count( 1 )
FROM
Score sc
WHERE
sc.c_id = c.c_id
)) “85-70”,((
SELECT
count( 1 )
FROM
Score sc
WHERE
sc.c_id = c.c_id
AND sc.s_score <= 70 AND sc.s_score > 60
)/(
SELECT
count( 1 )
FROM
Score sc
WHERE
sc.c_id = c.c_id
)) “70-60”,((
SELECT
count( 1 )
FROM
Score sc
WHERE
sc.c_id = c.c_id
AND sc.s_score <= 60 AND sc.s_score >= 0
)/(
SELECT
count( 1 )
FROM
Score sc
WHERE
sc.c_id = c.c_id
)) “60-0”
FROM
Course c
ORDER BY
c.c_id

1	-- 24、查询学生平均成绩及其名次

SET @rownum = 0;

SELECT
@rownum := @rownum + 1 AS rownum,
a.*
FROM
( SELECT s_id, AVG( s_score ) FROM Score GROUP BY s_id ORDER BY AVG( s_score ) DESC ) a

1	-- 25、查询各科成绩前三名的记录

SELECT
*
FROM
(
SELECT
st.s_id,
st.s_name,
st.s_sex,
st.s_birth,
sc.c_id,
sc.s_score
FROM
Student st,
Score sc
WHERE
st.s_id = sc.s_id
AND sc.c_id = ‘01’
ORDER BY
sc.s_score DESC
LIMIT 3
) a UNION ALL
SELECT
*
FROM
(
SELECT
st.s_id,
st.s_name,
st.s_sex,
st.s_birth,
sc.c_id,
sc.s_score
FROM
Student st,
Score sc
WHERE
st.s_id = sc.s_id
AND sc.c_id = ‘02’
ORDER BY
sc.s_score DESC
LIMIT 3
) b UNION ALL
SELECT
*
FROM
(
SELECT
st.s_id,
st.s_name,
st.s_sex,
st.s_birth,
sc.c_id,
sc.s_score
FROM
Student st,
Score sc
WHERE
st.s_id = sc.s_id
AND sc.c_id = ‘03’
ORDER BY
sc.s_score DESC
LIMIT 3
) c
```