Division in group by SQL table - sql

Question

I have the following SQL code
SELECT (COUNT(*) filter (WHERE has_a)) AS count_a,
COUNT(*) AS total_count,
process_date
FROM(SELECT process_date::date AS process_date,
(CASE WHEN (columnA > 0) THEN true ELSE false END) AS has_a
FROM my_table)
temptable
GROUP BY process_date
LIMIT 5;
which gives the following table
I want to create a column called percent_a with value (count_a/total_count)*100 grouped by process_date. E.g for row 1 the new column would have value 49.4 i.e. (1030/2085)*100.
I have tried
SELECT process_date,
((COUNT(*) filter (WHERE has_a))/COUNT(*) * 100) AS percent_a,
FROM(SELECT process_date::date AS process_date,
(CASE WHEN (columnA > 0) THEN true ELSE false END) AS has_a,
FROM my_table)
temptable
GROUP BY process_date
ORDER BY process_date DESC
LIMIT 1;
But this just gave 0s.
How can I create the column to display the % I want? I think something is happening with the GROUP BY but I don't know how to fix it.

Answer 1

It's because count returns an integer, you just need to cast it:
SELECT process_date,
(((COUNT(*) filter (WHERE has_balance))::DOUBLE PRECISION)/COUNT(*) * 100) AS percent_a,
FROM(SELECT process_date::date AS process_date,
(CASE WHEN (columnA > 0) THEN true ELSE false END) AS has_a,
FROM my_table)
temptable
GROUP BY process_date
ORDER BY process_date DESC
LIMIT 1;