How to count NULLS without using IS NULL in a WHERE clause

This came up the other day, someone wanted to know the percentage of NULL values in a column

Then I said "I bet you I can run that query without using a NULL in the WHERE clause, as a matter of fact, I can run that query without a WHERE clause at all!!"

The person then wanted to know more, so you know what that means.. it becomes a blog post  :-)

BTW, the PostgreSQL version of this blog post can be found here:  A quick and easy way to count the percentage of nulls without a where clause in PostgreSQL


To get started, first create this table and verify you have 9 rows

CREATE TABLE foo(bar int)
INSERT foo values(1),(null),(2),(3),(4),
 (null),(5),(6),(7)

SELECT * FROM foo

Here is what the output should be

bar
1
NULL
2
3
4
NULL
5
6
7

To get the NULL values and NON NULL values, you can do something like this

SELECT COUNT(*) as CountAll FROM foo WHERE bar IS NOT NULL
SELECT COUNT(*) as CountAll FROM foo WHERE bar IS  NULL

However, there is another way

Did you know that COUNT behaves differently if you use a column name compared to when you use *

Take a look

SELECT COUNT(*) as CountAll, 
  COUNT(bar) as CountColumn
FROM foo

If you ran that query, the result is the following

CountAll    CountColumn
----------- -----------
9           7

Warning: Null value is eliminated by an aggregate or other SET operation.


And did you notice the warning? That came from the count against the column

Let's see what Books On Line has to say


COUNT(*) returns the number of items in a group. This includes NULL values and duplicates.

COUNT(ALL expression) evaluates expression for each row in a group, and returns the number of nonnull values.

COUNT(DISTINCT expression) evaluates expression for each row in a group, and returns the number of unique, nonnull values.

This is indeed documented behavior

So now, lets change our query to return the percentage of non null values in the column

SELECT COUNT(*) as CountAll, 
  COUNT(bar) as CountColumn, 
  (COUNT(bar)*1.0/COUNT(*))*100 as PercentageOfNonNullValues 
FROM foo

Here is the output

CountAll    CountColumn percentageOfNonNullValues
----------- ----------- ---------------------------------------
9           7           77.777777777700

I just want to point out one thing,  the reason I have this * 1.0 in the query

(COUNT(bar)*1.0/COUNT(*))*100

I am doing * 1.0 here because count returns an integer, so you will end up with integer math and the PercentageOfNonNullValues would be 0 instead of 77.7777...


That's it for this short post.. hopefully you knew this, if not, then you know it now  :-)