Count Distinct field with group by

Question

Here I ma trying to do

SELECT b.sponso , b.source_id , Count(Distinct b.source_id) As nb_part
FROM buzzes b 
WHERE b.type = 3 AND b.sponso_id != 0
GROUP BY b.source_id , b.sponso 

And the output :

+------------+--------------+--------+
| Sponso     | source_id   | nb_part | 
+------------+--------------+--------+
| A          | 954711      | 1       | 
| A          | 587741      | 1       | 
| B          | 321447      | 1       | 
| B          | 350         | 1       |
+------------+--------------+--------+

And what I am expected to have

+------------+--------------+--------+
| Sponso     | source_id   | nb_part | 
+------------+--------------+--------+
| A          | 954711      | 2       | 
| B          | 321447      | 2       | 
+------------+--------------+--------+

I wanted to count the sum of nb_part and group by Sponso but I am stuck with my query

Answer 1

It seems you want the maximum source_id and the number of different source_id per sponso . "Per sponso" translates to GROUP BY sponso in SQL.

SELECT sponso, MAX(source_id) AS source_id, COUNT(DISTINCT source_id) AS nb_part
FROM buzzes
WHERE type = 3 AND sponso_id != 0
GROUP BY sponso
ORDER BY sponso;

Or maybe you want to aggregate over your query:

SELECT sponso, MAX(source_id) AS source_id, SUM(nb_part) AS nb_part
FROM
(
  SELECT b.sponso , b.source_id , Count(Distinct b.source_id) As nb_part
  FROM buzzes b 
  WHERE b.type = 3 AND b.sponso_id != 0
  GROUP BY b.source_id , b.sponso 
)
GROUP BY sponso
ORDER BY sponso;