SELECT SUM(NULLs) AS 'NULLS', SUM(NOTNULLs) AS 'NOTNULLs' FROM 
    (select count(*) AS 'NULLs', 0 as 'NOTNULLs' FROM us WHERE a is null
    UNION select 0 as 'NULLs', count(*) AS 'NOTNULLs' FROM us WHERE a is not null) AS x

这很糟糕，但它将返回一个带有2个cols的记录，指示null和非null的计数。

如果我理解正确，你想在一个列中计数所有NULL和所有NOT NULL…

如果是正确的:

SELECT count(*) FROM us WHERE a IS NULL 
UNION ALL
SELECT count(*) FROM us WHERE a IS NOT NULL

阅读评论后，编辑了完整的查询:]

SELECT COUNT(*), 'null_tally' AS narrative 
  FROM us 
 WHERE a IS NULL 
UNION
SELECT COUNT(*), 'not_null_tally' AS narrative 
  FROM us 
 WHERE a IS NOT NULL;

a为空的元素个数:

select count(a) from us where a is null;

a不为空的元素个数:

select count(a) from us where a is not null;

SELECT SUM(NULLs) AS 'NULLS', SUM(NOTNULLs) AS 'NOTNULLs' FROM 
    (select count(*) AS 'NULLs', 0 as 'NOTNULLs' FROM us WHERE a is null
    UNION select 0 as 'NULLs', count(*) AS 'NOTNULLs' FROM us WHERE a is not null) AS x

这很糟糕，但它将返回一个带有2个cols的记录，指示null和非null的计数。

对于非空值

select count(a)
from us

null值

select count(*)
from us

minus 

select count(a)
from us

因此

SELECT COUNT(A) NOT_NULLS
FROM US

UNION

SELECT COUNT(*) - COUNT(A) NULLS
FROM US

应该做这项工作

更好的是列标题是正确的。

SELECT COUNT(A) NOT_NULL, COUNT(*) - COUNT(A) NULLS
FROM US

在我的系统上进行的一些测试中，需要进行全表扫描。

我在postgres 10中创建了这个表，下面两种方法都有效:

从我们中选择count(*)

and

从我们中选择count(a为空)