有关系 s(sno,sname) c(cno,cname) sc(sno,cno,grade) 求 sql 查询语言
1. select sno from s,c,sc where s.sno=sc.sno and c.cno=sc.cno 2. select top 1 sno from sc order by grade desc 3. select count(*) from sc where grade>90 group by cno
S: sno, sname, age,sex,dept SC: sno, cno,grade C: cno,cname 用SQL实现以下功能:
(1)查询“张三”同学所学课程的成绩,列出sno,cno,grade.select sno,cno,grade from s,sc where sc.sno=sc.sno and sname=’张三’ (2)检索计算机系年龄20岁以上的同的学号和姓名 select sno,sname where dept=’计算机系’ and age>20 (3)统计…
CREATE TABLE S (SNO CHAR(4) NOT NULL, SNAME CHAR(20) NOT NULL, SADD CHAR(20), PRIMARY KEY(SNO));
创建一张表名为S的表.表中有三个字段,分别为:SNO,类型为字符型,长度为4个字符,不许值为空;SNAME,类型为字符型,长度为20个字符,不许值为空;SADD,类型为字符型,长度为20个字符,允许值为空.表的主键为SNO.
SQL问题: 供应商:S(SNO,SNAME,CITY,STATUS) 零件:P(PNO,PNAME,WEIGHT,COLOR,CITY)
用MySQL测试通过,尽量写标准SQL,如果是其他数据库,应该可以通过或稍加修改即可通过. 13. select t2.sname from sp t1 left join s as t2 on t1.sno=t2.sno left join p t3 on t1.pno=t3.pno where t3.color=’red’ group by t2.sname
设有学生表S(SNO,SN)(SNO为学生号,SN为姓名)
select s.sno,s.sn from s where not exists(select ” from sc where sc.sno=s.sno and sc.cno=’c1′)
结构与数据库求助!!S( Sno,Sn,sex,age,dept) C ( Cno,Cn,Ct); SC( Sno,Cno,SCORE)
1 update S set age=age+12 delete from S where sno=’S7’3 select sn,dept,age from S where age between 20 and 234 select distinct sno from SC5 select sno,score from SC where cno=’3′ order by score desc
求SQL(要求用循环嵌套) S(SNO,SNAME) SC(SNO,GRADE) 找出各个学生的平均成绩,输出学生姓名和平均成绩
select s.name,avg(sc.grade) from s, sc where s.sno=sc.sno group by s.sno按照学号分组,算出每组的平均值,然后和学生姓名对应.(如果有两个人名字相同,因为按照学号分组所以也不会算在一起,而是分为两条数据
sql解答有三个关系 S(sno,sname,age,sex,sdept) C(cno,cname,cdept,tname) tname
查询至少选修了2号课程和8号课程的学生姓名 select s.sname from s,sc where s.sno=sc.sno and sc.cno in (2,8) group by s.sname having count(*)>=2 查询张红的年龄 select age from s where sname=’张红’ 查询李明同学不及格的课程名称 select …
写出SQL语句,已知有基本表:学生:S(SNO,SNAME,SEX,DEPT,BIRTH)课程:C
1 select sno from sc where cno=’1′ 2 select sname from s where sno in ( select sno from sc where cno=’2′ ) 3 select sname from s where dept=’计算机系’ and datediff(year,birth,getdate())<19