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

Try

SELECT 
   SUM(ISNULL(a)) AS all_null,
   SUM(!ISNULL(a)) AS all_not_null
FROM us;

简单!

所有的答案要么是错误的，要么是非常过时的。

执行此查询的简单而正确的方法是使用COUNT_IF函数。

SELECT
  COUNT_IF(a IS NULL) AS nulls,
  COUNT_IF(a IS NOT NULL) AS not_nulls
FROM
  us

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。

正如我理解你的查询，你只需要运行这个脚本并获得Total Null,Total NotNull行，

select count(*) - count(a) as 'Null', count(a) as 'Not Null' from us;