用于获取基于同一表中2个条件的计数的单个SQL查询

我有这样的数据

现在，我需要一个查询来获取id计数，其中信息为“是”，id计数同时为“是”和“否”

单一查询：

SELECT COUNT(id) FROM table WHERE info = 'yes'

及

自

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您可以使用聚合和窗口函数完成此操作：

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

看。 结果:

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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

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您可以在不使用子查询的情况下执行此操作。这取决于观察到的不仅仅是以下ID的数量：

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

同样，赞成的人数也是如此。所以，两者都有的数字是id的数量减去no only的数量减去yes only的数量：

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;
    

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这应该可以做到这一点…它肯定不是高效或优雅的，但没有空值聚合警告

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你的预期结果是什么？既然Id having Yes是71,2,3,4,5,6,7，Id having and Yes and No两个变量只有31,4,6，它应该给我Id_as_Yes=7和Id_as_Yes_No=3这部作品！感谢@forpas没有工作，而且在600万数据上执行速度太慢

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

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

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;
    

Select
    (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]