Count Distinct with Answer side by side instead of underneath - postgresql

Question

Here is my query:
SELECT substring(date,1,10), count(distinct id),
CASE WHEN name IS NOT NULL THEN 1 ELSE 0 END
FROM table
WHERE (date >= '2015-09-01')
GROUP BY substring(date,1,10), CASE WHEN name IS NOT NULL THEN 1 ELSE 0 END
ORDER BY substring(date,1,10)
This is my result:
substring count case
2015-09-01 20472 0
2015-09-01 7 1
2015-09-02 20465 0
2015-09-02 470 1
What I want it to look like is this:
substring count count
2015-09-01 20472 7
2015-09-02 20465 470
Thank you!

Answer 1

With PostgreSQL 9.4 or newer, we can filter directly an aggregate with the new FILTER clause:
SELECT substring(date,1,10),
count(distinct id),
count(*) FILTER (WHERE name IS NOT NULL)
FROM table
WHERE (date >= '2015-09-01')
GROUP BY 1
ORDER BY 1

Answer 2

SELECT substring(date,1,10)
, count(distinct CASE WHEN name IS NOT NULL THEN id ELSE null END ) AS count1
, count(distinct CASE WHEN name IS NOT NULL THEN null ELSE id END ) AS count2
FROM event
WHERE (date >= '2015-09-01')
GROUP BY substring(date,1,10)
ORDER BY substring(date,1,10)
This gave me an answer like this: (which is exactly what I wanted so thank you so much)
substring count1 count2
2015-09-01 7 20472
2015-09-02 470 20465

Answer 3

Use case in count to get columns for some condition (name IS NOT NULL), like this:
SELECT substring(date,1,10)
, count(distinct CASE WHEN name IS NOT NULL THEN id ELSE null END ) AS count1
, count(distinct CASE WHEN name IS NOT NULL THEN null ELSE id END ) AS count2
FROM table
WHERE (date >= '2015-09-01')
GROUP BY substring(date,1,10)
ORDER BY substring(date,1,10)
you can also use subquery to create columns:
SELECT dt, Count(id1) count1, Count(distinct id2) count2
FROM (
SELECT distinct substring(date,1,10) AS dt
, CASE WHEN name IS NOT NULL THEN id ELSE null END AS id1
, CASE WHEN name IS NOT NULL THEN null ELSE id END AS id2,
FROM table
WHERE (date >= '2015-09-01')) d
GROUP BY dt
ORDER BY dt