根据上一篇文章建立的表,我们来做一些多表练习:
没建立表的可以点击此链接去建立练习用的表:
目录
12.查询被"Tom"和"Jerry"教的课程的最高分和最低分
15.查询课程编号为1且课程成绩在60分以上的学生的学号和姓名(子查询)
16. 查询平均成绩大于等于70的所有学生学号、姓名和平均成绩
19.查询每门课程的平均成绩,结果按照平均成绩降序排列,如果平均成绩相同,再按照课程编号升序排列
20.查询平均成绩大于60分的同学的学生编号和学生姓名和平均成绩
28.查询课程名称为"java",且分数低于60分的学生姓名和分数
进行student表和scores表的id相连接,course表和scores表的id相连接
SELECT
s.id sid,
s.`name` sname,
c.`name` cname,
sc.score
FROM
student s
LEFT JOIN scores sc ON s.id = sc.s_id
LEFT JOIN course c ON c.id = sc.c_id
WHERE
s.id = 1;

SELECT
c.id,
c.`name`,
AVG( sc.score ),
max( sc.score )
FROM
course c
LEFT JOIN scores sc ON c.id = sc.c_id
GROUP BY
c.id,
c.`name`;

SELECT
s.id,
s.`name`,
c.`name` cname,
sc.score
FROM
student s
LEFT JOIN scores sc ON sc.s_id = s.id
LEFT JOIN course c ON c.id = sc.c_id
WHERE
s.`name` LIKE '张%';

SELECT
t.id,
t.NAME,
c.id,
c.NAME,
r.score
FROM
(
SELECT
s.id,
s.NAME,(
SELECT
max( score )
FROM
scores r
WHERE
r.s_id = s.id
) score
FROM
student s
) t
LEFT JOIN scores r ON r.s_id = t.id
AND r.score = t.score
LEFT JOIN course c ON r.c_id = c.id;

SELECT
*
FROM
student s
WHERE
id IN (
SELECT DISTINCT
r.s_id
FROM
(
SELECT
c.id,
c.NAME,
max( score ) score
FROM
student s
LEFT JOIN scores r ON r.s_id = s.id
LEFT JOIN course c ON c.id = r.c_id
GROUP BY
c.id,
c.NAME
) t
LEFT JOIN scores r ON r.c_id = t.id
AND t.score = r.score
);

SELECT
s.id,
s.NAME sname,
sc.score,
c.NAME
FROM
student s
LEFT JOIN scores sc ON s.id = sc.s_id
LEFT JOIN course c ON sc.c_id = c.id
WHERE
s.NAME LIKE '%张%'
OR s.NAME LIKE '%李%';

SELECT
*
FROM
student
WHERE
id IN (
SELECT
sc.s_id
FROM
scores sc
GROUP BY
sc.s_id
HAVING
avg( sc.score ) >= 70
);

SELECT
s.id,
s.NAME,
sum( sc.score ) score
FROM
student s
LEFT JOIN scores sc ON s.id = sc.s_id
GROUP BY
s.id,
s.NAME
ORDER BY
score DESC,
s.id ASC;

SELECT
c.NAME,
max( sc.score ),
min( sc.score ),
avg( sc.score )
FROM
course c
LEFT JOIN scores sc ON c.id = sc.c_id
WHERE
c.NAME = '数学';

SELECT
c.id,
c.NAME,
avg( sc.score ) score
FROM
course c
LEFT JOIN scores sc ON c.id = sc.c_id
GROUP BY
c.id,
c.NAME
ORDER BY
score DESC;

SELECT
t.id,
t.NAME,
c.id cid,
c.NAME cname,
avg( r.score )
FROM
teacher t
LEFT JOIN course c ON t.id = c.t_id
LEFT JOIN scores r ON r.c_id = c.id
GROUP BY
t.id,
t.NAME,
c.id,
c.NAME;

SELECT
t.id,
t.NAME,
c.id cid,
c.NAME cname,
max( r.score ),
min( r.score )
FROM
teacher t
LEFT JOIN course c ON t.id = c.t_id
LEFT JOIN scores r ON r.c_id = c.id
GROUP BY
t.id,
t.NAME,
c.id,
c.NAME
HAVING
t.NAME IN ( 'Tom', 'Jerry' );

SELECT
t.id,
t.sname,
r.c_id,
c.NAME,
t.score
FROM
(
SELECT
s.id,
s.NAME sname,
max( r.score ) score
FROM
student s
LEFT JOIN scores r ON r.s_id = s.id
GROUP BY
s.id,
s.NAME
) t
LEFT JOIN scores r ON r.s_id = t.id
AND r.score = t.score
LEFT JOIN course c ON r.c_id = c.id;

SELECT
s.id,
s.NAME,
c.id,
c.NAME,
r.score
FROM
student s
LEFT JOIN scores r ON s.id = r.s_id
LEFT JOIN course c ON c.id = r.c_id;

SELECT
s.*,
r.*
FROM
student s
LEFT JOIN scores r ON s.id = r.s_id
WHERE
r.c_id = 1
AND r.score > 60

SELECT
s.id,
s.NAME,
t.score
FROM
student s
LEFT JOIN ( SELECT r.s_id, avg( r.score ) score FROM scores r GROUP BY r.s_id ) t ON s.id = t.s_id
WHERE
t.score >= 70;

SELECT
*
FROM
student s
WHERE
id IN ( SELECT r.s_id FROM scores r GROUP BY r.s_id HAVING min( r.score ) < 60 );

SELECT
c.id,
c.NAME,
count(*)
FROM
course c
LEFT JOIN scores r ON c.id = r.c_id
GROUP BY
c.id,
c.NAME;

SELECT
c.id,
c.NAME,
avg( score ) score
FROM
course c
LEFT JOIN scores r ON c.id = r.c_id
GROUP BY
c.id,
c.NAME
ORDER BY
score DESC,
c.id ASC;

SELECT
s.id,
s.NAME sname,
avg( r.score ) score
FROM
student s
LEFT JOIN scores r ON r.s_id = s.id
LEFT JOIN course c ON c.id = r.c_id
GROUP BY
s.id,
s.NAME
HAVING
score > 65;

SELECT
s.id,
s.NAME,
s.gender
FROM
student s
LEFT JOIN scores r ON s.id = r.s_id
WHERE
r.score > 80
GROUP BY
s.id,
s.NAME,
s.gender
HAVING
count(*) = 1;

SELECT
s.id,
s.NAME,
s.gender
FROM
student s
LEFT JOIN scores r ON s.id = r.s_id
GROUP BY
s.id,
s.NAME,
s.gender
HAVING
count(*) = 3;

SELECT
*
FROM
course c
WHERE
id IN (
SELECT
r.c_id
FROM
scores r
GROUP BY
r.c_id
HAVING
min( r.score ) < 60
);

SELECT
s.id,
s.NAME
FROM
student s
LEFT JOIN scores r ON s.id = r.s_id
GROUP BY
s.id,
s.NAME
HAVING
count(*) >= 4;

SELECT
*
FROM
student
WHERE
id IN (
SELECT
r.s_id
FROM
scores r
GROUP BY
r.s_id
HAVING
count(*) != 5
);

SELECT
s.id,
s.NAME,
count(*) number
FROM
student s
LEFT JOIN scores r ON s.id = r.s_id
GROUP BY
s.id,
s.NAME
HAVING
number = ( SELECT count(*) FROM course );

SELECT
s.id,
s.NAME,
count(*) number
FROM
student s
LEFT JOIN scores r ON s.id = r.s_id
GROUP BY
s.id,
s.NAME;

SELECT
s.id,
s.NAME,
r.score
FROM
student s
LEFT JOIN scores r ON s.id = r.s_id
LEFT JOIN course c ON r.c_id = c.id
WHERE
c.NAME = 'java'
AND r.score < 60;

SELECT
s.id,
s.NAME
FROM
student s
LEFT JOIN scores r ON r.s_id = s.id
LEFT JOIN course c ON c.id = r.c_id
LEFT JOIN teacher t ON t.id = c.t_id
WHERE
t.NAME = 'Tom';

SELECT
*
FROM
student
WHERE
id NOT IN (
SELECT DISTINCT
s.id
FROM
student s
LEFT JOIN scores r ON r.s_id = s.id
LEFT JOIN course c ON c.id = r.c_id
LEFT JOIN teacher t ON t.id = c.t_id
WHERE
t.NAME = 'Tom'
)
文章目录一、概述简介原理模块二、配置Mysql使用版本环境要求1.操作系统2.mysql要求三、配置canal-server离线下载在线下载上传解压修改配置单机配置集群配置分库分表配置1.修改全局配置2.实例配置垂直分库水平分库3.修改group-instance.xml4.启动监听四、配置canal-adapter1修改启动配置2配置映射文件3启动ES数据同步查询所有订阅同步数据同步开关启动4.验证五、配置canal-admin一、概述简介canal是Alibaba旗下的一款开源项目,Java开发。基于数据库增量日志解析,提供增量数据订阅&消费。Git地址:https://github.co
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我是Ruby的新手。我安装了DataMapper并且正在尝试安装dm-mysql-adapter-1.0.2gem。但是当我尝试安装时,出现以下错误。我正在使用ubuntu操作系统。vinoth@vinoth-laptop:~/Downloads$geminstalldm-mysql-adapter-1.0.2----with-mysql-lib=/usr/lib/mysql----with-mysql-conf=/usr/bin/mysqlWARNING:Installingto~/.gemsince/home/vinoth/gemsand/home/vinoth/gems/bina
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