这里有两种解决方案:

Select count(columnname) as countofNotNulls, count(isnull(columnname,1))-count(columnname) AS Countofnulls from table name

OR

Select count(columnname) as countofNotNulls, count(*)-count(columnname) AS Countofnulls from table name

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)

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;

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

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;

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