Is it possible to combine Group by, Having and Sum?

Question

I have a table:

------------------------
|id|p_id|desired|earned|
------------------------
|1 | 1  |  5    |  7   |
|2 | 1  |  15   |  0   |
|3 | 1  |  10   |  0   |
|4 | 2  |  2    |  3   |
|5 | 2  |  2    |  3   |
|6 | 2  |  2    |  3   |
------------------------

I need to make some calculations, and try to make it in one not really complex request, otherwise I know how to calculate it with numbers of requests. I need resulted table like following:

---------------------------------------------------------
|p_id|total_earned|    AVG   |      Count     |  SUM    |
|    |            | (desired)|(if earned != 0)|(desired)|
---------------------------------------------------------
|  1 |      7     |     10   |       1        |    30   |
|  2 |      9     |      2   |       3        |    6    |
---------------------------------------------------------

I build so far:

SELECT p_id, SUM(earned), AVG(desired), Sum(desired) 
FROM table GROUP BY p_id

But I can't figure out how to calculate the number of grouped records with conditions. I can get this number with HAVING but in separated request.

I almost sure what SQL should have this power.

Answer 1

You can use CASE expression for this.

Try this,

SELECT p_id
    ,SUM(earned) AS total_earned
    ,AVG(desired) AS avg_desired
    ,COUNT(CASE WHEN Earned!=0 THEN 1 END) AS earned_count
    ,SUM(desired) AS sum_desired
FROM table
GROUP BY p_id;

Answer 2

A shorter alternative to CASE is

SELECT p_id,
    SUM(earned) AS total_earned,
    AVG(desired) AS average_desired,
    COUNT(earned != 0 OR NULL) AS earned_count,
    SUM(desired) AS sum_desired
FROM table GROUP BY p_id;

because NULLs are not counted.