SELECT
    ALL_VALUES
    ,COUNT(ALL_VALUES)
FROM(
        SELECT 
        NVL2(A,'NOT NULL','NULL') AS ALL_VALUES 
        ,NVL(A,0)
        FROM US
)
GROUP BY ALL_VALUES

如果是mysql，你可以尝试这样做。

select 
   (select count(*) from TABLENAME WHERE a = 'null') as total_null, 
   (select count(*) from TABLENAME WHERE a != 'null') as total_not_null
FROM TABLENAME

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

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

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

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;