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!

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;

下面是一个在Oracle上运行的快速而简单的版本:

select sum(case a when null then 1 else 0) "Null values",
       sum(case a when null then 0 else 1) "Non-null values"
from us

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

我有一个类似的问题:要计算所有不同的值，空值也算作1。简单的计数在这种情况下不起作用，因为它不考虑空值。

下面是一个适用于SQL的代码片段，不涉及选择新值。 基本上，执行distinct后，还使用row_number()函数返回新列(n)中的行号，然后对该列执行计数:

SELECT COUNT(n)
FROM (
    SELECT *, row_number() OVER (ORDER BY [MyColumn] ASC) n
    FROM (
        SELECT DISTINCT [MyColumn]
                    FROM [MyTable]
        ) items  
) distinctItems