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

以防你想把它记录在一条记录里:

select 
  (select count(*) from tbl where colName is null) Nulls,
  (select count(*) from tbl where colName is not null) NonNulls 

;-)

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;

Try

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

简单!

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。

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