让我提前为这个问题的简单性道歉(我听到Jeff的播客，以及他担心问题的质量将被"降低")，但我被困住了。我正在使用AquaData打我的Informix数据库。 MS SQL和Informix SQL之间有一些古怪的细微差别。无论如何，我试图做一个简单的嵌套表达式，但它讨厌我。

select 
  score,
  count(*) students,
  count(finished) finished,
  count(finished) / count(*)students   
--  round((count(finished) / count(*)students),2) 
from now_calc 
group by score
order by score

简单除法表达式的行返回完成人员的百分比，这正是我想要的...我只需要将结果舍入到2位即可。注释行(-)不起作用。我尝试了所有可能想到的变化。

*我不打算同时使用第5行和第6行

抱歉，我应该提到now_calc是一个临时表，并且字段名称实际上是" students"和" finished"。我之所以这样命名，是因为我要直接将这些结果吐到Excel中，并且我希望将字段名加倍作为列标题。因此，我了解我们在说什么，基于此，我通过删除(*)使其工作：

select 
  score,
  count(students) students,
  count(finished) finished,
  round((count(finished) / count(students) * 100),2) perc
from now_calc 
group by score
order by score

我将整个查询包括在内，这对于其他任何人都可能更有意义。从学习的角度来看，重要的是要注意在" finished"字段上进行计数的唯一原因是因为Case语句根据Case语句的值将值设为1或者为null。如果该案例陈述不存在，则对"完成"计数将产生与对"学生"计数完全相同的结果。

--count of cohort members and total count of all students (for reference)
select 
  cohort_yr, 
  count (*) id,
  (select count (*) id from prog_enr_rec where cohort_yr is not null and prog = 'UNDG' and cohort_yr >=1998) grand
from prog_enr_rec
where cohort_yr is not null 
and prog = 'UNDG'
and cohort_yr >=1998
group by cohort_yr
order by cohort_yr;

--cohort members from all years for population
select 
  id,  
  cohort_yr,
  cl,
  enr_date,
  prog
from prog_enr_rec
where cohort_yr is not null 
and prog = 'UNDG'
and cohort_yr >=1998
order by cohort_yr
into temp pop with no log;

--which in population are still attending (726)
select 
  pop.id, 
  'Y' fin 
from pop, stu_acad_rec
where pop.id = stu_acad_rec.id 
and pop.prog = stu_acad_rec.prog
and sess = 'FA'
and yr = 2008
and reg_hrs > 0
and stu_acad_rec.cl[1,1] <> 'P'
into temp att with no log;

--which in population graduated with either A or B deg (702)
select 
  pop.id,
  'Y' fin
from pop, ed_rec
where pop.id = ed_rec.id
and pop.prog = ed_rec.prog
and ed_rec.sch_id = 10 
and (ed_rec.deg_earn[1,1] = 'B'
or  (ed_rec.deg_earn[1,1] = 'A'
and pop.id not in (select pop.id 
           from pop, ed_rec
           where pop.id = ed_rec.id
           and pop.prog = ed_rec.prog
           and ed_rec.deg_earn[1,1] = 'B'
           and ed_rec.sch_id = 10)))
into temp grad with no log;

--combine all those that either graduated or are still attending
select * from att
union
select * from grad
into temp all_fin with no log;

--ACT scores for all students in population who have a score (inner join to eliminate null values)
--score > 50 eliminates people who have data entry errors - SAT scores in ACT field
--2270
select 
  pop.id,
  max (exam_rec.score5) score
from pop, exam_rec
where pop.id = exam_rec.id
and ctgry = 'ACT'
and score5 > 0 
and score5 < 50
group by pop.id
into temp pop_score with no log;

select 
  pop.id students,
  Case when all_fin.fin = 'Y' then 1 else null end finished,
  pop_score.score
from pop, pop_score, outer all_fin
where pop.id = all_fin.id 
and pop.id = pop_score.id
into temp now_calc with no log;

select 
  score,
  count(students) students,
  count(finished) finished,
  round((count(finished) / count(students) * 100),2) perc
from now_calc 
group by score
order by score

谢谢！

解决方案

SELECT
        score,
        count(*) students,
        count(finished) finished,
        count(finished) / count(*) AS something_other_than_students,   
        round((count(finished) / count(*)),2) AS rounded_value
    FROM now_calc 
    GROUP BY score
    ORDER BY score;

请注意，输出列名称" students"被重复使用，这也使我们感到困惑。我使用的AS是可选的。

我现在已经针对IDS正式验证了语法，并且可以使用：

Black JL: sqlcmd -Ffixsep -d stores -xf xx.sql | sed 's/        //g'
+ create temp table now_calc(finished CHAR(1), score INTEGER, name CHAR(10) PRIMARY KEY);
+ insert into now_calc values(null, 23, 'a');
+ insert into now_calc values('y',  23, 'b');
+ insert into now_calc values('y',  23, 'h');
+ insert into now_calc values('y',  23, 'i');
+ insert into now_calc values('y',  23, 'j');
+ insert into now_calc values('y',  43, 'c');
+ insert into now_calc values(null, 23, 'd');
+ insert into now_calc values('y',  43, 'e');
+ insert into now_calc values(null, 23, 'f');
+ insert into now_calc values(null, 43, 'g');
+ SELECT
        score,
        count(*) students,
        count(finished) finished,
        count(finished) / count(*) AS something_other_than_students,
        round((count(finished) / count(*)),2) AS rounded_value
    FROM now_calc
    GROUP BY score
    ORDER BY score;
 23|       7|       4| 5.71428571428571E-01|      0.57
 43|       3|       2| 6.66666666666667E-01|      0.67
Black JL:

我让'finished'取空值是因为'count(finished)/ count(*)'不返回1的唯一原因是'finished'接受nulls-虽然不是很好的表设计。然后我将得分为23的7行放入小数位数(然后将得分为43的一行更改为第二个具有小数位数的数字)。