Take precedence on a specific value from a table - sql

Question

For each person's distinct record that has a toyota,
only take toyota and filter out that person's other cars
else bring all cars.
The actual script will not match my logic above. I was trying to simplify my question by using random names and car brands, but the objective was the same since I wanted to get a specific address code and filter out the rest if it did exist for other vendor names (see below). Thank you, GMB.
GPMEM.dbo.PM00200 a -- Vendor Master
LEFT JOIN GPMEM.dbo.PM30200 b -- Historical/Paid Transactions
ON a.VENDORID = b.VENDORID
LEFT JOIN GPMEM.dbo.PM20000 c -- Open/Posted Transactions
ON a.VENDORID = c.VENDORID
LEFT JOIN (
SELECT d.*,
rank() over(
partition by d.VENDORID
order by case when d.ADRSCODE = 'ACH' THEN 0 ELSE 1 END
)rn
FROM GPMEM.dbo.PM00300 d
) d -- Vendor Address Master
ON a.VENDORID = d.VENDORID
WHERE
d.rn = 1

Answer 1

You can use window functions:
select colA, colB
from (
select
t.*,
rank() over(
partition by colA
order by case when colB = 'Toyota' then 0 else 1 end
) rn
from mytable t
) t
where rn = 1
The trick likes in the order by clause in the over() clause of window function rank(): if a person has a Toyota, it will be ranked first, and their (possible) other cars will be ranked second. If it has no Toyota, all their car will be ranked first.

Answer 2

You can do this with filtering logic:
select t.*
from t
where t.colb = 'toyota' or
not exists (select 1 from t t2 where t2.cola = t.cola and t2.colb = 'toyota');
If I were to use window functions for this, I would simply count the toyotas:
select t.*
from (select t.*,
sum(case when colb = 'toyota' then 1 else 0 end) over (partition by cola) as num_toyotas
from t
) t
where colb = 'toyota' or num_toyotas = 0;