按同一个连接表过滤时计算所有连接表行

Question

So I'm querying a table districts, that has one to many relationship with the positions table. I want to select the districts that have the positions with particular name AND count ALL districts positions at the same time. Can I do this?

With the current code

 SELECT districts.*, COUNT(DISTINCT(positions.id)) as positions_count FROM "districts"
  LEFT JOIN positions ON positions.id = districts.position_id
 WHERE ("positions"."name" IN ($1, $2)) GROUP BY districts.id ORDER BY positions_count desc, "districts"."name" ASC

If I have 20 positions in some district but only 2 are filtered, positions_count eq 2 as well, I want it to be 20

I've tried to join twice on the same table with an alias but this gives me the same result

SELECT districts.*, COUNT(DISTINCT(positions_to_count.id)) as positions_count FROM "districts"
  LEFT JOIN positions ON positions.id = districts.position_id
  LEFT JOIN positions AS positions_to_count ON positions_to_count.id = districts.position_id WHERE ("positions"."name" IN ($1, $2)) GROUP BY districts.id ORDER BY positions_count desc, "districts"."name" ASC

Answer 1

I think you just want conditional aggregation:

SELECT d.*, COUNT(DISTINCT p.id) as positions_count,
       COUNT(DISTINCT CASE WHEN p.name IN ($1, $2) THEN p.id END) as positions_name
FROM "districts" d LEFT JOIN
     positions p 
     ON p.id = d.position_id 
GROUP BY d.id
ORDER BY positions_count desc, d."name" ASC;

Note: If you do not have duplicates for a given district, then there is no need to use COUNT(DISTINCT) , just COUNT() suffices.

Answer 2

Use the filtered positions to get the districts that you want in the results and then join all the positions:

SELECT d.*, 
  COUNT(DISTINCT p.id) as positions_count
FROM (
  SELECT DISTINCT d.* 
  FROM districts AS d INNER JOIN positions AS p 
  ON p.id = d.position_id
  WHERE p.name IN ($1, $2)
) AS d INNER JOIN positions AS p
ON p.id = d.position_id
GROUP BY d.id 
ORDER BY positions_count DESC, d.name ASC

I'm not sure about DISTINCT p.id . You can also try without DISTINCT .