1楼：羊驷
(1) select t.tname,t.titlefrom t inner join c on t.t=c.twhere c.**ame="高等数学"
(2) select dlookup("**ame","c","c='" & sc.c & "'") as **ame
from s left join sc on s.s = sc.swhere s.sname="zhang" and sc.s is null
(3) select c.**ame, c.cfrom c inner join sc on c.
c = sc.cwhere sc.s="s2" or sc.
s="s9"
(4) select s, sname
from s
where age>20 and ***="男"
急！请大侠们指点，一道数据库的题。写出关系代数表达式~~非常感谢！ 50
2楼：匿名用户
关系代数的符号不好表达呀
两者理解起来都一样的……
很简单的
3楼：杨力闻
1.use database_name
select s.sno,c.credit from course c join elective e
on c.con=e.con join student s
on s.sno=e.sno
where c.con='c06'
2..use database_name
select s.sname,c.sname from course c join elective e
on c.con=e.con join student s
on s.sno=e.sno
where c.con='c06'
3.select s.sname,c.sname from course c join elective e
on c.con=e.con join student s
on s.sno=e.sno
where c.**ame='english'
4..select s.sno from course c join elective e
on c.con=e.con join student s
on s.sno=e.sno
where c.**o in('c02','c06')
5...select s.sno from course c left join elective e
on c.con=e.con join student s
on s.sno=e.sno
where c.**o in('c02','c06')
6...select s.sno from course c left join elective e
on c.con=e.con join student s
on s.sno=e.sno
where c.**o not in('c02','c06')
7.select sname from student
8.select s.sno from course c left join elective e
on c.con=e.con join student s
on s.sno=e.sno
where s.sno like['s08'%]
呵呵，就这些了
求助 这道数据库 关系代数表达式的题目 的答案
4楼：巍科软件
--1.检索“吴迪”老师讲授的课程号和课程名
select **o ,**ame ,teacher from c where teacher ='吴迪'
--2.检索所有女同学和年龄小于20岁的男同学
select * from s where
s.***='男' and age <20 or s .*** ='女'
--3.检索至少选修“吴迪”老师讲授课程中一门课的学生姓名
select sname from s where sno in (select sno from sc where **o in (select **o from c where teacher ='吴迪' ))
--4.检索“李波”同学不学的课程的课程名
select **ame from c where **o in (select **o from sc where sno not in (select sno from s where sname ='李波'))
--5.检索至少选修两门课程的学生的姓名
select sname from s where sno in (select sno from (select count (sno) as 's**umber',sno from sc group by sno) scs where s**umber >=2)
--6.检索未被选修的课程的课程名
select **ame from c where **o not in (select **o from sc)
--7.检索选修全部课程的学生的学号
select sno from s where sno in (select sno from (select count (sno) as 's**umber',sno from sc group by sno) scs where s**umber = (select count (distinct c .**ame) from c ))
--8.检索选修了“吴迪”老师讲授课程的学生的学号
select sno from s where sno in (select sno from sc where **o in (select **o from c where teacher ='吴迪'))
--9.检索选修了“吴迪”老师讲授课程且成绩高于85的学生的学号sele
select sno from s where dept >85 and sno in (select sno from sc where **o in (select **o from c where teacher ='吴迪'))
--10.检索“王虎”同学所学课程的课程号
select **o from sc where sno = (select sno from s where sname ='王虎')
--11.检索选修了c01和c02两门课程的学生的姓名
select sname from s where sno in ( select sc.sno from sc where sc.sno in(select sc.
sno from sc where sc.**o = (select c.**o from c where c.
**ame ='c01'))and sc.**o=(select c.**o from c where c.
**ame ='c02'))
--12.检索未选修课程的学生的学号
select sno from s where sno not in (select sno from sc ) select * from sc
一道数据库题目，谁能帮我解答下，急！！！
5楼：匿名用户
π是投影,该运算从表中选出指定的属性组成一个新表,记为:πa(r).其中a是属性名(即列名)表,r是表名
σ是选择,该运算按给定的条件,从表中选出满足条件的行形成一个新表,作为一个新表.如:σf(r).其中f为条件表达式,r为表名
1. π姓名,家庭地址σ 职务='科长'(职工)
2. π姓名,家庭地址σ 部门.部门名称='办公室'∧职工.职务='科长'(部门.部门编号∞职工.所属部门编号)
3. π姓名,家庭地址σ部门名称='财务科'∧职务= '科长'(部门.部门编号∞职工.所属部门编号)
4. del from 职工 where 职工号='3016'
5. update 保健 set 健康状况='一般' where 职工号='3016'
6. create view 视图名 as selcet * from 保健 where 健康状况='差'
数据库相关问题，用关系代数表达式表示
6楼：匿名用户
sql语句如下：
1.select * from student where sname='李政'
2.这题见鬼了，没有年龄字段，更没有出生日期等信息，写什么？
3.select sname,sno from student where exists(select 1 from class where clsname='计算机01' and clsno=student.clsno)
4.select clsname from class where exists(select 1 from student where sname='张山' and clsno=class.clsno)
求帮忙做一道数据库题。对高手来说很简单的，帮帮忙，谢谢。
7楼：匿名用户
是要用关系代数表达式吗？全部直接sql语句写完不行吗？
8楼：
1.πcid,**ame(σe68a84e8a2ad62616964757a686964616f31333330343161 teacher='liu'(c))
select cid,**ame from c where teacher = 'liu'
2.πsid,sname(σ age>23∧***='男'(s))
select sid,sname from s where age>23 and ***='男'
3.π**ame,teachre(σ sid = 's3' (c×sc))
select c.cid, **ame,teachre from sc s, c c
where sid = 's3' and s.cid = c.cid
4.πsname(σ *** = '女' ∧teacher='liu' (s×c×sc))
select sname from s s
where exists(select 1 from sc sc,c c where c.teacher='liu' and sc.cid = c.
cid and s.sid=sc.sid) and s.
*** ='女'
5.πcid(sc）－πcid(σ sname='wang'（s×sc）)
6.π1(σ1=4∧2≠5(sc×sc）)
7.πcid,**ame(c×(πsid,cid(sc)÷πsid(s)))
8.πsid,cid(sc)÷ πcid(σ teacher = 'liu'(sc×c)）
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