mysql

/*create table teacher(
    tno varchar(10) primary key,-- 教师编号
    tname varchar(20) not null,-- 教师姓名
    sex char(2) default ‘男‘,-- 性别
    age date,-- 年龄
    sal int,-- 工资
    dname varchar(30) -- 所属系别
)*/
/*insert into teacher values(‘1001‘,‘张三‘,‘男‘,‘1987-08-08‘,3500,‘计算机‘);
insert into teacher values(‘1002‘,‘李四‘,‘男‘,‘1983-06-10‘,4500,‘计算机‘)*/

#insert into teacher values(‘1003‘,‘李四‘,‘女‘,‘1984-06-10‘,4500,‘计算机‘)
#insert into teacher values(‘1004‘,‘王五‘,‘女‘,‘1988-06-10‘,4000,‘计算机‘)
#insert into teacher values(‘1005‘,‘啊啊‘,‘女‘,‘1988-06-10‘,4000,‘物理系‘)

#insert into teacher values(‘1008‘,‘王二小‘,‘女‘,‘1988-06-10‘,4000,‘计算机‘)
#排序  先按照工资排序   如果工资一样那么再按照年龄排序    asc代表升序   desc 代表降序
#select * from teacher ORDER BY sal DESC,age desc


#查询工资最高的三个人

#select * from teacher ORDER BY sal desc limit 3

#别名
#select tname ‘姓名‘,sal ‘工资‘ from teacher

#查询工资在4000到5000之间
#select * from teacher where sal>4000 and sal<5000

#select * from teacher where sal BETWEEN 4000 and 5000


#select * from teacher where age is not null


#查找出教师工资大于4000的  或者 小于 3000的

#select * from teacher where sal>4000 or sal<3000

#查找出计算机系或者物理系的教师工资大于3000的
#select * from teacher where (dname=‘计算机‘ or dname=‘物理系‘) and sal>3000


#select * from teacher where dname in(‘计算机‘,‘物理系‘) and sal>3000

#select * from teacher where (dname=‘计算机‘ and sal>3000) or (dname=‘物理系‘ and sal>3000)
#查询姓王的老师
#select * from teacher where sal like ‘4%‘


/*create VIEW v2
as
select * from teacher where tno in(select tno from tc where course=‘java‘)*/




#select * from v2





/*create VIEW v1
as
select tname,sal from teacher*/

/*create view v3(t,s)
as
select tname,sal from teacher*/


#select * from v3


/*alter VIEW v1
as
select * from teacher*/


#select * from v1

#insert into v1(tno,tname) values(‘1008‘,‘abc‘)


/*create view v4
AS
select avg(sal) FROM teacher*/

#select * from v4


#update from v4 set avg(sal)=3000




#select * from v1 



#select avg(sal) from teacher

#insert into teacher VALUES(‘1100‘,‘aa‘,‘男‘,‘1999-09-09‘,2000,‘数学‘)



#insert into teacher select * from teacher



#select * from teacher where tno=‘1100‘


#alter table teacher
#add index ix(tno)




#select avg(sal) into @a from teacher ;
#select @a


/*create PROCEDURE fun()
BEGIN
    set @i=0;
    set @sum=0;
    while @i<101 DO
        set    @sum=@sum+@i;
        set    @i=@i+1;
        end while;
    select @sum;
end*/


/*create PROCEDURE fun1()
BEGIN
    DECLARE sun int;   
    DECLARE i int;
    set sun=0;
    set i=0;
    while i<101 DO
        set    sun=sun+i;
        set    i=i+1;
        end while;
    select sun;
end*/


#call fun1()

#平均工资大于4000显示工资高，否则显示工资低


/*create PROCEDURE fun2()
BEGIN
        select avg(sal) into @a from teacher;
    IF(@a>4000) THEN
        select ‘工资高‘;
    ELSE
        select ‘工资低‘;
    end if;
end
*/

#call fun2()

#创建一个带输入参数的存储过程，查看指定教师编号的教师信息


/*create PROCEDURE fun3(in id VARCHAR(20))
BEGIN
select * from teacher where tno=id;
end*/

#call fun3(‘1002‘)


#创建一个带有输出参数的存储过程，返回教师表的最高工资


/*create PROCEDURE fun4(out m INT)
BEGIN

    select MAX(sal) into m FROM teacher;
    
END*/



#call fun4(@m);

#select @m


#创建一个函数返回指定系的教师平均工资


/*create FUNCTION f1(d varchar(20))
RETURNS INT
BEGIN
        select avg(sal) into @avg from teacher;
        return @avg;
END*/

#SELECT f1(‘计算机‘)

#select avg(sal) from teacher


/*create PROCEDURE p1(in d VARCHAR(20))
BEGIN
    select * from teacher where dname=d;
end*/

#select * from (call p1(‘计算机‘))


#select STRCMP(‘dbc‘,‘bc‘)









/*select tname,
case
 when age>‘1990‘ then ‘年轻‘
    when age<‘1985‘ then ‘老年‘
    else ‘中年‘ END as ‘age‘
from Teacher*/
    

/*create table bank(
id VARCHAR(10) PRIMARY key,
bname VARCHAR(10),
money INT
)*/


/*create table info(
carid varchar(10),
transDate datetime,
type VARCHAR(10),
money int
)*/


#update bank set money=money-200 where id=‘1001‘;

#insert into info VALUES(1001,CURRENT_TIMESTAMP(),‘支取‘,‘200‘)


/*create TRIGGER t1
AFTER update on bank
for each ROW
BEGIN
    insert into info VALUES(‘1001‘,CURRENT_TIMESTAMP(),‘支取‘,200);
end
*/


/*create TRIGGER t2
AFTER update on bank
for each ROW
BEGIN
            if(new.money<0) THEN
                insert into a values(1,2);
            end if;*/

        /*if(old.money-new.money>0) THEN
            insert into info VALUES(old.id,‘支取‘,old.money-new.money);
        else 
            insert into info VALUES(old.id,CURRENT_TIMESTAMP(),‘存款‘,new.money-old.money);
        end if;*/
#end


update bank set money=money-10000 where id=‘1001‘;