你再试一下，这次应该可以了

select
s.studentName as 学生姓名,
sum(c.late) as 迟到,
sum(c.leave) as 早退
from
student s
inner join check c on s.studentID = c.studentID
group by
s.studentName

[此贴子已经被作者于2007-3-6 12:19:35编辑过]

好的，谢谢这里的热心朋友，我先试试！

哈哈，＂棉花糖ONE＂是正解，要用 left join才能求出所有的学生；我稍做修改如下：
select s.studentName,sum(c.late),sum(c.leave) from student s left join check c on c.studentID=s.studentID group by s.studentName