Sql 根据where语句从子查询中选择多个结果计数？

因此，我从子查询中进行选择，并根据使用where语句创建的列中的值是否包含值来显示这些行的计数

我想创建一个查询，根据不同的where语句返回几个计数

我的表有两列：ID1和abc2

select count(*) from
(
select `id1`, count(distinct(`abc2`)) as total_abc
from TABLE
where `id1` != 'Unknown' and `abc2` != 'NULL'
group by `id1`
order by total_abc desc
)
where total_abc = 1




select count(*) from
(
select `id1`, count(distinct(`abc2`)) as total_abc
from TABLE
where `id1` != 'Unknown' and `abc2` != 'NULL'
group by `id1`
order by total_abc desc
)
where total_abc = 2





select count(*) from
(
select `id1`, count(distinct(`abc2`)) as total_abc
from TABLE
where `id1` != 'Unknown' and `abc2` != 'NULL'
group by `id1`
order by total_abc desc
)
where total_abc = 3

所需的结果只是将以下几个查询组合在一起：

count1 | count2 | count3
__________________________
123    | 222     | 34567

---

您可以使用两个级别的聚合：

select
    count(case when num_vals = 1 then 1 end) as val_1,
    count(case when num_vals = 2 then 1 end) as val_2,
    count(case when num_vals = 3 then 1 end) as val_3
from
(
    select id1, count(distinct abc2) as num_vals
    from TABLE
    where id1 <> 'Unknown' and abc2 <> 'NULL'
    group by id1
) t;

您似乎想要列中的值。我更喜欢成排排列：

select num_vals, count(*)
from (select id1, count(distinct abc2) as num_vals
      from TABLE
      where id1 <> 'Unknown' and abc2 <> 'NULL'
      group by id1
     ) t
group by num_vals
order by num_vals;

---

因为您的子查询是完全相同的，所以您可以这样做

select 
SUM(CASE WHEN total_abc = 1 THEN 1 ELSE 0 END) Count1
SUM(CASE WHEN total_abc = 2 THEN 1 ELSE 0 END) Count2
SUM(CASE WHEN total_abc = 3 THEN 1 ELSE 0 END) Count3
 from
(
select `id1`, count(distinct(`abc2`)) as total_abc
from TABLE
where `id1` != 'Unknown' and `abc2` != 'NULL'
group by `id1`
order by total_abc desc
)
where total_abc IN (1,2,3)

---

请用您正在使用的数据库标记您的问题。嗨，如果我想对所有值执行此操作，该怎么办？所以1-无穷大？如果你想让它以列的形式返回，可能很难做到，除非你是透视的。我建议看看上面Gordon的答案，他使用分组来实现相同的结果，但是成排的。