随着数字化时代的飞速发展,数据库技术,特别是MySQL,已经成为IT领域中不可或缺的一环。从简单的数据存储到复杂的数据分析,从个人博客到大型企业的核心系统,MySQL都在背后默默地发挥着作用。而在这不断演变的过程中,我们对于MySQL的了解和掌握也需要与时俱进。
虚拟表,和普通表一样使用
创建:
create view 视图名 as 查询语句 ;
CREATE VIEW aa as SELECT s.*,sc.cid,sc.score FROM t_mysql_student s, t_mysql_score sc WHERE s.sid = sc.sid;
查看:
show create view 视图名;
修改:
create or replace view 视图名
as 查询语句
1.查询" 01 “课程比” 02 "课程成绩高的学生的信息及课程分数
SELECT
s.*,
( CASE WHEN t1.cid = ‘01’ THEN t1.score END ) 语文,
( CASE WHEN t2.cid = ‘02’ THEN t2.score END ) 数学
FROM
t_mysql_student s,
( SELECT * FROM t_mysql_score WHERE cid = ‘01’ ) t1,
( SELECT * FROM t_mysql_score WHERE cid = ‘02’ ) t2
WHERE
s.sid = t1.sid
AND t1.sid = t2.sid
AND t1.score > t2.score
02)查询同时存在" 01 “课程和” 02 "课程的情况
SELECT
s.*,
( CASE WHEN t1.cid = ‘01’ THEN t1.score END ) 语文,
( CASE WHEN t2.cid = ‘02’ THEN t2.score END ) 数学
FROM
t_mysql_student s,
( SELECT * FROM t_mysql_score WHERE cid = ‘01’ ) t1,
( SELECT * FROM t_mysql_score WHERE cid = ‘02’ ) t2
WHERE
s.sid = t1.sid
AND t1.sid = t2.sid
03)查询存在" 01 “课程但可能不存在” 02 "课程的情况(不存在时显示为 null )
SELECT * from
( SELECT * FROM t_mysql_score WHERE cid = ‘01’ ) t1
LEFT JOIN
( SELECT * FROM t_mysql_score WHERE cid = ‘02’ ) t2
on t1.sid=t2.sid
04)查询不存在" 01 “课程但存在” 02 "课程的情况
SELECT
*
FROM
t_mysql_student s,
t_mysql_score sc
WHERE
s.sid = sc.sid
AND s.sid NOT IN ( SELECT sid FROM t_mysql_score WHERE cid = ‘01’ )
AND sc.cid = ‘02’
05)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
s.sid,
s.sname,
ROUND( AVG( sc.score ) ) 分数
FROM
t_mysql_student s
LEFT JOIN t_mysql_score sc ON s.1sid = sc.sid
GROUP BY
s.sid,
s.sname
HAVING
分数 >= 60
06)查询在t_mysql_score表存在成绩的学生信息
SELECT
s.sid,
s.sname
FROM
t_mysql_student s
INNER JOIN t_mysql_score sc ON s.sid = sc.sid
GROUP BY
s.sid,
s.sname
07)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(没成绩的显示为 null )
SELECT
s.sid,
s.sname,
COUNT( sc.cid ),
SUM( sc.score )
FROM
t_mysql_score sc,
t_mysql_student s
WHERE
sc.sid = s.sid
GROUP BY
s.sid,
s.sname
08)查询「李」姓老师的数量
select count(*) from t_mysql_teacher where tname like ‘李%’