您可以使用聚集和窗口函数执行此操作：

SELECT DISTINCT 
       SUM(MAX(CASE WHEN info = 'yes' THEN 1 ELSE 0 END)) OVER () id_as_yes,
       COUNT(CASE WHEN COUNT(DISTINCT info) = 2 THEN 1 END) OVER () id_as_yes_no
FROM tablename
GROUP BY id

请参见demo。
结果：

> id_as_yes | id_as_yes_no
> --------: | -----------:
>         7 |            3

您需要条件聚合。

Select id, 
       Count(case when info = 'y' then 1 end) as y_count,
       Count(case when info = 'y' and has_n = 1 then 1 end) as yn_count
  From (SELECT id, info,
               Max(case when info = 'no' then 1 end) over (partirion by id) as has_n
         From your_table) t

您可以在没有子查询的情况下执行此操作。这依赖于以下观察结果：仅为“no”的id的数量为：

count(distinct id) - count(distinct case when info = 'yes' then id end)

对于肯定的数目也是如此，所以，两者都有的数目等于id的数目减去否定的数目再减去肯定的数目：

select count(distinct case when info = 'yes' then id end) as num_yeses,
       (count(distinct id) -
        (count(distinct id) - count(distinct case when info = 'yes' then id end)) -
        (count(distinct id) - count(distinct case when info = 'no' then id end))
       )
from t;

这应该可以解决问题......它肯定不是高效或优雅的，但没有空值聚合警告
dbFiddle link

Select
    (select count(distinct id) from mytest where info = 'yes') as yeses
    ,(select count(distinct id) from mytest where info = 'no' and id in (select distinct id from mytest where info = 'yes' )) as [yes and no]