SQL查询问题，高手帮帮忙。。。

[风影萧诺]

有张学生出勤表 stu_table
字段：stu_id，stu_type (注：stu_type: 1 = 出勤，2 = 病假/事假，3 = 迟到/早退 4 = 旷课)

用一条sql返回以下格式结果:

应出勤 实出勤 出勤率 病假/事假 迟到/早退 旷课

[--小F--]

select
  stu_id,
  count(1) as 应出勤,
  sum(case wtu_type when 1 then 1 else 0 end) as [实出勤],
  ltrim(sum(case wtu_type when 1 then 1 else 0 end)*100.0/count(1))+'%' as  出勤率,
  sum(case wtu_type when 2 then 1 else 0 end) as [病假/事假],
  sum(case wtu_type when 3 then 1 else 0 end) as [迟到/早退],
  sum(case wtu_type when 4 then 1 else 0 end) as [旷课]
group by
   stu_id

[--小F--]

select
  stu_id,
  count(1) as 应出勤,
  sum(case wtu_type when 1 then 1 else 0 end) as [实出勤],
  ltrim(sum(case wtu_type when 1 then 1 else 0 end)*100.0/count(1))+'%' as  出勤率,
  sum(case wtu_type when 2 then 1 else 0 end) as [病假/事假],
  sum(case wtu_type when 3 then 1 else 0 end) as [迟到/早退],
  sum(case wtu_type when 4 then 1 else 0 end) as [旷课]
from
  tb
group by
   stu_id

--如果不是统计每个学生的 就这样
select
  count(1) as 应出勤,
  sum(case wtu_type when 1 then 1 else 0 end) as [实出勤],
  ltrim(sum(case wtu_type when 1 then 1 else 0 end)*100.0/count(1))+'%' as  出勤率,
  sum(case wtu_type when 2 then 1 else 0 end) as [病假/事假],
  sum(case wtu_type when 3 then 1 else 0 end) as [迟到/早退],
  sum(case wtu_type when 4 then 1 else 0 end) as [旷课]
from
  tb

[风影萧诺]

很专业。。。 谢谢了。。。
就是统计整个班级的。。、。。