SQL 舍入不带小数的整数

Question

我有一个看似简单的问题。我只想显示没有小数位的百分比，并且总加起来为 100。这里有一个小片段：

create table genderTable
(person varchar(10),
isMale varchar(5))

insert into genderTable values ('Mary', 'false')
insert into genderTable values ('Frank', 'true')
insert into genderTable values ('Bill', 'true')
insert into genderTable values ('Jessie', 'false')
insert into genderTable values ('Sue', 'false')
insert into genderTable values ('Beth', 'false')
insert into genderTable values ('Kris', 'false')

declare @total as int
set @total = 7
select
CASE isMale
   WHEN 'True' THEN 'Male'
   ELSE 'Female'
END as Gender,
CASE 
WHEN @total > 0 THEN ROUND((count(isMale) * 100 / @total), 0)
ELSE 0
END as GenderPercent
from genderTable
group by  isMale

总数加起来是 99%，而不是 100%。我尝试了各种四舍五入，但我要么得到一位小数，要么得到 99%。有什么帮助吗？请记住，在另一个例子中，我必须对有两个以上值的种族做同样的事情，所以从 100 中减去可能是行不通的......

Answer 1

对于你的特殊问题，当我这样做时，我得到的数字加起来是 100：

select isMale, count(*), sum(count(*)) over (),
       round(100.0 * count(*) / sum(count(*)) over (), 0)
from genderTable t
group by isMale;

您的实现的实际问题是 SQL Server 执行整数运算。因此，表达式ROUND((count(isMale) * 100 / @total), 0) 正在使用整数除法进行计算——在实现round() 之前取比率的floor()。

有很多方法可以做你想做的事。它们在 SQL Server 2012+ 中比在早期版本中更容易实现：

select isMale,
       round(100.0 * cnt / tot, 0) as p,
       (case when seqnum = 1
             then 100 - sum(round(100.0 * cnt / tot, 0)) over (order by seqnum desc rows between unbounded preceding and 1 preceding)
             else round(100.0 * cnt / tot, 0) 
        end) as p_tot_100
from (select isMale, count(*)*1.0 as cnt, sum(1.0*count(*)) over () as tot,
             row_number() over (order by isMale) as seqnum
      from genderTable t
      group by isMale
     ) t;

这个想法是将所有其他行的舍入版本相加，然后从 100 中减去其中一行。

编辑：

这些的“整数”版本：

select isMale, count(*), sum(count(*)) over (),
       cast(round(100.0 * count(*) / sum(count(*)) over (), 0) as int)
from genderTable t
group by isMale;

和：

select isMale,
       round(100.0 * cnt / tot, 0) as p,
       cast((case when seqnum = 1
                  then 100 - sum(round(100.0 * cnt / tot, 0)) over (order by seqnum desc rows between unbounded preceding and 1 preceding)
                  else round(100.0 * cnt / tot, 0) 
             end) as int) as p_tot_100
from (select isMale, count(*)*1.0 as cnt, sum(1.0*count(*)) over () as tot,
             row_number() over (order by isMale) as seqnum
      from genderTable t
      group by isMale
     ) t;

我很确定即使使用浮点表示，小整数也能准确表示，因此round() 生成诸如 29.99999999999997 而不是 30 之类的值不会有问题。

Answer 2

我刚刚看到上面的答案。干得好，戈登！

这是答案的类似版本：

SELECT CASE isMale
    WHEN 'True'
        THEN 'Male'
    ELSE 'Female'
    END AS Gender
   ,CAST(left(round(count(isMale) * 100.00 /(select count(*) from gendertable), 0),2) AS VARCHAR) + '%' AS GenderPercent
FROM genderTable
GROUP BY isMale