为了提供另一种选择，Postgres 9.4+允许对聚合应用FILTER:

SELECT
  COUNT(*) FILTER (WHERE a IS NULL) count_nulls,
  COUNT(*) FILTER (WHERE a IS NOT NULL) count_not_nulls
FROM us;

SQLFiddle: http://sqlfiddle.com/ # !17/80a24/5

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;

在我的例子中，我想要多列之间的“空分布”:

SELECT
       (CASE WHEN a IS NULL THEN 'NULL' ELSE 'NOT-NULL' END) AS a_null,
       (CASE WHEN b IS NULL THEN 'NULL' ELSE 'NOT-NULL' END) AS b_null,
       (CASE WHEN c IS NULL THEN 'NULL' ELSE 'NOT-NULL' END) AS c_null,
       ...
       count(*)
FROM us
GROUP BY 1, 2, 3,...
ORDER BY 1, 2, 3,...

根据“……它可以很容易地扩展到更多的列，只要需要

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。

我通常用这个技巧

select sum(case when a is null then 0 else 1 end) as count_notnull,
       sum(case when a is null then 1 else 0 end) as count_null
from tab
group by a

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。