各位帮个忙,万分感谢

2024-11-17 14:51:41
推荐回答(5个)
回答1:

create table worker(
wokerid int,
workername char(8),
sex char(2),
borndata datetime,
partymember bit,
job datetime,
departid int);

create table depart(
departid int,
departname char(10));

create table salary(
workerid int,
data datetime,
pay decimal(6,1));

1.
create table salary(
workerid int,
data datetime,
pay decimal(6,1),
PRIMARY KEY (workerid,data),
FOREIGN KEY (workerid) REFERENCES worker(workerid));

2.
select workerid,localtime-borntime
from worker

3.
create view aaa as
select departid,count(workerid)
from worker
where partymember=0
group by departid

4.
select workerid,workername,departname,pay
from worker,depart,salary
where worker.workerid=salary.workerid
and worker.departid=depart.departid
and data='2007-02'
order by departname desc

5.
update salary
set pay=pay*1.05
where pay<(select avg(pay) from salary)

6.
create index i_woker on worker(departid asc,borndata desc)

7.

先到这里,刚做完数据库试验,还比较清晰,加分的话再作剩下的好了。

回答2:

一般情况下,这么点分没有人做的

回答3:

你这要求太高了,你简直希望别人帮你设计一套完成的程序,我们只是解决问题,自己在编写程序的过程中遇到问题我们可以帮你,你这不是帮,是要我们帮你做事呢!!

回答4:

爱莫能助啊

回答5:

给点劳务费还差不多,其他费用就不收你的了,500左右