计算每个类别中的记录

Question

I have two tables:

ci_categories
with fields cat_id , cat_name , cat_slug etc.

ci_albums
with fields cat_id , album_id , album_name etc.

I want to join these two tables, display category listing, and COUNT albums in each category. Like:

• Category 1 - 20 Albums

• Category 2 - 25 Albums

How is this done properly in PostgreSQL?

Answer 1

The following query should get you what you want:

SELECT c.cat_id, COUNT(a.album_id)
FROM ci_categories AS c
LEFT JOIN ci_albums AS a ON c.cat_id = a.cat_id
GROUP BY c.cat_id

We use a LEFT JOIN here due to the possibility of categories which contain no albums. If you don't care if those results show up, just change this to an INNER JOIN .

We use the aggregate COUNT(a.album_id) function to count the number of records. By grouping by the c.cat_id , we make sure this count is done only over records of the same type. If there is no album for that category, a.album_id will be NULL, and so COUNT will not include it in the total count.

Answer 2

select c.cat_name, 
       count(distinct a.album_id) as albums
from ci_categories c
left join ci_albums a on a.cat_id = c.cat_id
group by c.cat_name

Answer 3

SELECT ci_categories.cat_name, COUNT(*) as count

FROM ci_categories JOIN ci_albums ON ci_categories.cat_id = ci_albums.cat_id

GROUP BY ci_categories.cat_name;

The other answers are somewhat equal with some slight differences, notably the DISTINCT in the COUNT and the aliasing of the tables.

EDIT / WARNING

Well, my answer is actually wrong, due to the fact that a JOIN will also include a categories if there are no albums in that category. That category is then joined to NULL values, which are counted by COUNT(*) anyways. Leaving this one here as an example of a common mistake.

Answer 4

If you are satisfied with cat_id and the count of albums in the result, you don't need the table ci_categories in your query at all. Use this simpler and faster query:

SELECT cat_id, count(*) AS albums
FROM   ci_albums
GROUP  BY 1;

Assuming proper table definitions without duplicate entries.

This does not includes categories with no albums at all. Only if you need those - too, or if you need additional information from the table ci_categories itself - only then you need to join to that table. But since we are counting everything , it is faster to aggregate first and join later:

SELECT c.cat_id, c.cat_name, COALESCE(a.albums, 0) AS albums
FROM   ci_categories c
LEFT   JOIN (
   SELECT cat_id, count(*) AS albums
   FROM   ci_albums
   GROUP  BY 1
   ) a USING (cat_id);

LEFT JOIN if you want categories with 0 albums, too. Else, just JOIN .
COALESCE is only needed if you need 0 instead of NULL in the result.
Test performance with EXPLAIN ANALYZE if you care.