SQL - Take Corresponding Value Where MIN(Date)

Question

I have a dataset with warehouse locations, dates, and inventory levels for given warehouse at a given date.

How do I structure a SQL query where I end up with a list of distinct locations and their inventory levels at the earliest date per warehouse? I'm thinking something like...

SELECT warehouse_id, (inventory_amt where min(date) is TRUE) as inventory

FROM inventory_table

GROUP BY warehouse_id

Answer 1

I like using row_number() and top (1) with ties for this:

SELECT TOP (1) WITH TIES it.*
FROM inventory_table it
ORDER BY ROW_NUMBER() OVER (PARTITION BY warehouse_id ORDER BY date DESC);

You can also express this as:

select it.*
from inventory_table it
where it.date = (select max(it2.date)
                 from inventory_table it2
                 where it2.warehouse_id = it.warehouse_id
                );

Answer 2

try something like this

select warehouse_id,....
from  inventory_table t1
inner join (
select warehouse_id, min(date) from  inventory_table group py warehouse_id
) min
on t1.warehouse_id=min.warehouse_id

Answer 3

This worked for me:

proc sql;
CREATE TABLE inventory_by_warehouse AS
SELECT
    a.warehouse_id
    ,a.inventory_amt
FROM inventory_table as a
WHERE a.date = 
    (SELECT min(date)
    FROM inventory_table
    WHERE warehouse_id = a.warehouse_id);
quit;