我有一张桌子
create table us
(
a number
);
现在我有如下数据:
a
1
2
3
4
null
null
null
8
9
现在我需要一个查询来计算列a中的空值和非空值
我有一张桌子
create table us
(
a number
);
现在我有如下数据:
a
1
2
3
4
null
null
null
8
9
现在我需要一个查询来计算列a中的空值和非空值
当前回答
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。
其他回答
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
Try
SELECT
SUM(ISNULL(a)) AS all_null,
SUM(!ISNULL(a)) AS all_not_null
FROM us;
简单!
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;
SELECT SUM(NULLs) AS 'NULLS', SUM(NOTNULLs) AS 'NOTNULLs' FROM
(select count(*) AS 'NULLs', 0 as 'NOTNULLs' FROM us WHERE a is null
UNION select 0 as 'NULLs', count(*) AS 'NOTNULLs' FROM us WHERE a is not null) AS x
这很糟糕,但它将返回一个带有2个cols的记录,指示null和非null的计数。
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