Counting null and non-null values in a single query

Question

I have a table

create table us
(
 a number
);

Now I have data like:

a
1
2
3
4
null
null
null
8
9

Now I need

Answer 1

Number of elements where a is null:

select count(a) from us where a is null;

Number of elements where a is not null:

select count(a) from us where a is not null;

Answer 2

Just to provide yet another alternative, Postgres 9.4+ allows applying a FILTER to aggregates:

SELECT
  COUNT(*) FILTER (WHERE a IS NULL) count_nulls,
  COUNT(*) FILTER (WHERE a IS NOT NULL) count_not_nulls
FROM us;

SQLFiddle: http://sqlfiddle.com/#!17/80a24/5

Answer 3

SELECT
    ALL_VALUES
    ,COUNT(ALL_VALUES)
FROM(
        SELECT 
        NVL2(A,'NOT NULL','NULL') AS ALL_VALUES 
        ,NVL(A,0)
        FROM US
)
GROUP BY ALL_VALUES

Answer 4

If I understood correctly you want to count all NULL and all NOT NULL in a column...

If that is correct:

SELECT count(*) FROM us WHERE a IS NULL 
UNION ALL
SELECT count(*) FROM us WHERE a IS NOT NULL

Edited to have the full query, after reading the comments :]

---

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;

Answer 5

if its mysql, you can try something like this.

select 
   (select count(*) from TABLENAME WHERE a = 'null') as total_null, 
   (select count(*) from TABLENAME WHERE a != 'null') as total_not_null
FROM TABLENAME

Answer 6

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

It's fugly, but it will return a single record with 2 cols indicating the count of nulls vs non nulls.