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。

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)

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

Try

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

简单!

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