推荐回答(5个)
select 学号,
max(case 课程号 when ‘101’ then 分数 else 0end) 101,
max(case 课程号 when '102' then 分数 else 0 end) 102,
max(case 课程号 when '103' then 分数 else 0 end) 103
from Biao
group by 学号
得到以下记录集
学号 101 102 103
1 100 90 80
2 90 80 100
3 80 100 90
然后在这基础上
select 学号 ,
max(case when 101=max(101) then 学号 else null end) 101,
max(case when 102=max(102) then 学号 else 0 end) 102,
max(case when 103=max(103) then 学号 else 0 end) 103
from
(
select 学号,
max(case 课程号 when ‘101’ then 分数 else 0end) 101,
max(case 课程号 when '102' then 分数 else 0 end) 102,
max(case 课程号 when '103' then 分数 else 0 end) 103
from Biao
group by 学号
) a
group by 学号
这会得到一个记录集
学号 101 102 103
1 1 null null
2 null null 2
3 null 3 null
然后最后一步操作就简单了
sum就好
这是比较麻烦的方法,
有简单的.
先查出每门课程的最高分,然后再和 biao 进行匹配一下即可,如:
select a.*
from biao a
where exists (select 1
from (select t.课程号, max(t.分数) 分数
from biao t
group by t.课程号) s
where a.课程号 = s.课程号
and a.分数 = s.分数)
select 学号,max(101)as 学分 from biao
UNION
select 学号,max(102)as 学分 from biao
UNION
select 学号,max(102)as 学分 from biao
试试,没有测试。
select 学号 from (select 分数,学号,课程号 from biao order by 分数) group by 课程号
楼主试试我这个
select 学号 from biao where (课程号,分数) in (select 课程号,max(分数) from biao group by 课程号)
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