Making rank() ignore null cases - sql

Question

I'm trying to figure out a way to make the rank() window function "skip" in its counting some rows with null values in a specific column. Pretty much, what I want is to count how many paid transactions there are, for each client, before each transaction/row.
I tried using case when inside the rank() and I got something similar to the results I expect, but still not quite what I need.
+-------------------------------------------------------+
| What I need |
+-------------+------+----------+-----------------------+
| CLIENT | CODE | PAYMENT | PAID_PURCHASES_SO_FAR |
| A | 341 | 17/09/21 | 0 |
| A | 342 | 18/09/21 | 1 |
| A | 343 | (null) | 2 |
| A | 344 | 18/09/21 | 2 |
| A | 345 | 19/09/21 | 3 |
| A | 346 | 19/09/21 | 4 |
| A | 347 | (null) | 5 |
| A | 348 | 24/09/21 | 5 |
| B | 855 | (null) | 0 |
| B | 856 | 20/09/21 | 0 |
| B | 857 | (null) | 1 |
+-------------+------+----------+-----------------------+
-+------------------------------------------------------+
| What I got |
-+------------+------+----------+-----------------------+
| CLIENT | CODE | PAYMENT | PAID_PURCHASES_SO_FAR |
| A | 341 | 17/09/21 | 0 |
| A | 342 | 18/09/22 | 1 |
| A | 343 | (null) | (null) |
| A | 344 | 18/09/22 | 2 |
| A | 345 | 19/09/22 | 3 |
| A | 346 | 19/09/21 | 4 |
| A | 347 | (null) | (null) |
| A | 348 | 24/09/21 | 5 |
| B | 855 | (null) | (null) |
| B | 856 | 20/09/21 | 0 |
| B | 857 | (null) | (null) |
-+------------+------+----------+-----------------------+
In a single image: comparison
And here my code:
SELECT
CLIENT
, CODE
, PAYMENT
, CASE WHEN PAYMENT IS NOT NULL THEN DENSE_RANK() OVER(PARTITION BY CLIENT, (CASE WHEN PAYMENT IS NOT NULL THEN 1 ELSE 0 END) ORDER BY CODE) - 1 END NUMBER_OF_PURCHASES_SO_FAR
FROM FOO.BAR
Note: The CODE column may be used as time reference. E.g. code = 750 came before code = 751, and so on.
Any help would be appreciated.
Thanks in advance.

Answer 1

You can use conditional aggregation combined with a window frame, as in:
select *, coalesce(sum(case when payment is null then 0 else 1 end)
over(partition by client order by code
rows between unbounded preceding and 1 preceding), 0)
as ppsf
from t
order by client, code
Result:
client code payment ppsf
------- ----- --------- ----
A 341 17/09/21 0
A 342 18/09/21 1
A 343 null 2
A 344 18/09/21 2
A 345 19/09/21 3
A 346 19/09/21 4
A 347 null 5
A 348 24/09/21 5
B 855 null 0
B 856 20/09/21 0
B 857 null 1
See running example at db<>fiddle.

Answer 2

This is it:
SELECT
"CLIENT"
, "CODE"
, "PAYMENT"
, rank() over(partition by "CLIENT"
order by COALESCE("PAYMENT",'01/01/70') )
FROM Table1
http://sqlfiddle.com/#!15/f941a/14