求一个Sql的写法（可加分），兄弟们帮帮忙

[fishzone]

有这么一个学生答案表: studentanswer
student_id test_id question_id model_answer student_answer
0000000001 00001 1 A A
0000000002 00001 1 A B
.
.
.
.
想在我想统计同一次考试的每一个题目有多少人答对了。用一个Sql写完，应该怎么写？
如果用n(n为question_id的个数)条sql语句，可以这样写：
select count(student_answer) as countanswer from studentanswer where test_id=1 and question_id=1 and model_answer=student_answer
不想这样写，是因为如果n很大的话，这样会很慢。

[foolishtiger]

用存储过程会快很多.就SQL语句的写法来说,你的思路是对的.

[karma]

select question_id, count(student_id)
from studentanswer
where model_answer=student_answer
group by question_id

[fishzone]

谢谢你的关注，我本来想这样写：
select count(student_answer) as countanswer from studentanswer where test_id=00001 group by question_id
这样可以返回每一个题目有多少个人做答，但是，怎样才可以返回每一道题有多少个（（正确））答案呢？

[karma]

the answer I provided is exactly what you want

[lisbjing]

to karma：
你明白他的意思嗎？不知道他要什麼?
select question_id, count(student_id)
from studentanswer
where model_answer=student_answer
group by question_id

[fishzone]

karma(无为):
你把 where model_answer=student_answer 放在了group by 之前，如果没有正确答案的话，它将返回不了这个question_id 下的 recordset而不是countanswer=0

[shkingstargroup]

同一次考试的每一个题目有多少人答对了
select question_id, count(student_id)
from studentanswer
where model_answer=student_answer and test_id=00001
group by question_id

[karma]

select question_id, sum( case when model_answer=student_answer then 1 else 0 end )
from studentanswer
group by question_id

[fishzone]

karma(无为):
试一下，谢谢。

[fishzone]

问题解决，谢谢各位。特别是 karma(无为)　：）
但是我用的数据库是access,所以我把你的语句改成：
select question_id, sum(iif(model_answer=student_answer,1,0))
from studentanswer
group by question_id

case wher ... then ...else 在这里不行，它好像是用在Sql server中的