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

如果你正在使用MS Sql Server…

SELECT COUNT(0) AS 'Null_ColumnA_Records',
(
    SELECT COUNT(0)
    FROM your_table
    WHERE ColumnA IS NOT NULL
) AS 'NOT_Null_ColumnA_Records'
FROM your_table
WHERE ColumnA IS NULL;

我不建议你这么做……但这里你有它(在同一张表中的结果)

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;

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

使用ISNULL嵌入函数。

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;