(2) select * from ( select a.xh,c.skcmc,c.dkcmc from Xjjbxx a join dm_bj b on a.bjdm = b.bjdm and b.zydm = '0101' join ztob c on b.zydm = c.zydm ) a left join zkcj b on a.xh = b.xh and (a.skcmc = b.kcmc or a.dkcmc = b.kcmc) where b.kcmc is null
select z.* from dm_bj d inner join Xjjbxx x on d.bjdm=x.bjdm inner join ztob z on z.zydm=d.zydm where d.zydm='0101' and not exists(select * from zkcy where kcmc=z.skcmc)
1、就是选择专业课程成绩,为了得到课程对应用中的相关信息才用那样的处理 2、以第一的关系为基础 select a.* from ztob a join dm_bj b on a.zydm=b.zydm join xjjbxx c on b.bjdm=c.bjdm where c.xh='0101' and not exists (select 1 from zkcj where kcmc=a.skcmc and xh=c.xh)
(一): select a.课程名称,a.成绩,a.学分,课程对应表.* from (select 学生原始成绩表.*,班级信息表.班级对应的专业的代码 from 学生原始成绩表 a,学生信息表 c,班级信息表 d where 学生原始成绩表.学号=学生信息表.学号 and 学生信息表.在班级代码=班级信息表.在班级代码 ) a left outer join 课程对应表 b on a.课程名称=课程对应表.原始课程名称 and a.班级对应的专业的代码=课程对应表.专业代码 order by 课程名称查出所有学生原始成绩信息的所有课程名称,成绩,学分,以及课程对应表信息 条件是学生原始成绩表的课程名称和课程对应表的课程名称相同,并且班级信息表的专业和课程对应表的专业相同
(二): select skcmc as 原始课程名称 from ztob where zydm='0101' and xh not in(select xh from Zkcj a,dm_bj b,kjjbxx c where a.xh=c.xh and c.bjdm=b.bjdm and b.zydm='0101')
select * from
(
select a.xh,c.skcmc,c.dkcmc
from Xjjbxx a join dm_bj b
on a.bjdm = b.bjdm and b.zydm = '0101'
join ztob c on b.zydm = c.zydm
) a left join zkcj b
on a.xh = b.xh and (a.skcmc = b.kcmc or a.dkcmc = b.kcmc)
where b.kcmc is null
from dm_bj d inner join Xjjbxx x on d.bjdm=x.bjdm inner join ztob z on z.zydm=d.zydm
where d.zydm='0101' and not exists(select * from zkcy where kcmc=z.skcmc)
2、以第一的关系为基础
select a.* from ztob a join dm_bj b on a.zydm=b.zydm
join xjjbxx c on b.bjdm=c.bjdm where c.xh='0101' and not exists
(select 1 from zkcj where kcmc=a.skcmc and xh=c.xh)
select a.课程名称,a.成绩,a.学分,课程对应表.*
from (select 学生原始成绩表.*,班级信息表.班级对应的专业的代码 from 学生原始成绩表 a,学生信息表 c,班级信息表 d where 学生原始成绩表.学号=学生信息表.学号 and 学生信息表.在班级代码=班级信息表.在班级代码 ) a
left outer join 课程对应表 b on a.课程名称=课程对应表.原始课程名称 and a.班级对应的专业的代码=课程对应表.专业代码
order by 课程名称查出所有学生原始成绩信息的所有课程名称,成绩,学分,以及课程对应表信息
条件是学生原始成绩表的课程名称和课程对应表的课程名称相同,并且班级信息表的专业和课程对应表的专业相同
select skcmc as 原始课程名称 from ztob where zydm='0101' and xh not in(select xh from Zkcj a,dm_bj b,kjjbxx c where a.xh=c.xh and c.bjdm=b.bjdm and b.zydm='0101')