用一条SQL语句 查询出每门课都大于80分的学生姓名，表( #test）如下：
Name Course Mark
张三 语文 81
张三 数学 75
李四 语文 76
李四 数学 90
王五 英语 100
王五 语文 81
create table #test
(
Name varchar(10),
Course varchar(10),
Mark float
)
insert into #test
select '张三', '语文', 81 union
select '张三', '数学', 75 union
select '李四', '语文', 76 union
select '李四', '数学', 90 union
select '王五', '英语', 100 union
select '王五', '语文', 81
方法A select distinct Name
from #test A
where not exists(select Course
from #test
where Mark < 80 and Name = A.Name)
方法B
select * from #test a where mark > 80 and (select count(*) from #test where name=a.name)= (select count(*) from #test where name=a.name and mark > 80)
方法C
select distinct name from #test a where not exists(select * from #test where a.name=name and mark<80) （我认为这种较好）
方法D
select distinct name from #test where
name not in (
select name from #test where mark<=80
)
方法E
select name,min(mark) from #test group by name having min(mark)>80
-----66666
DECLARE @test table ( Name varchar(10), Course varchar(10), Mark float)
insert into @test select '张三 ', '语文 ', 81
union select '张三 ', '数学 ', 75
union select '李四 ', '语文 ', 76
union select '李四 ', '数学 ', 90
union select '王五 ', '英语 ', 100
union select '王五 ', '语文 ', 81
SELECT NAME FROM @test GROUP BY name HAVING count(*)=count(case when mark>=80 then 1 else null end) 或者
SELECT NAME FROM @test GROUP BY name HAVING count(*)=sum(case when mark>=80 then 1 else 0 end)
---方法F
select name from #test group by name having min(mark)>80