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)

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

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 

;-)

在阿尔贝托的基础上，我添加了汇总。

 SELECT [Narrative] = CASE 
 WHEN [Narrative] IS NULL THEN 'count_total' ELSE    [Narrative] END
,[Count]=SUM([Count]) FROM (SELECT COUNT(*) [Count], 'count_nulls' AS [Narrative]  
FROM [CrmDW].[CRM].[User]  
WHERE [EmployeeID] IS NULL 
UNION
SELECT COUNT(*), 'count_not_nulls ' AS narrative 
FROM [CrmDW].[CRM].[User] 
WHERE [EmployeeID] IS NOT NULL) S 
GROUP BY [Narrative] WITH CUBE;

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

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