3

Let's say this is the sample date coming from a join of 2 tables. Database is Postgres 9.6

id  product_id  invoice_id  amount       date
1    PROD1       INV01       2          01-01-2018
2    PROD2       INV02       3          01-01-2018
3    PROD1       INV01       2          05-01-2018
4    PROD1       INV03       1          05-01-2018
5    PROD2       INV02       3          08-01-2018
6    PROD2       INV04       4          08-01-2018

I want to know if it's possible in a optimized way to:

  1. Get all the PRODx with their respective INVx which have the latest date, but per product_id. Please note that records unused from a day may be reported to a new one. This means:
id  product_id  invoice_id  amount       date
3    PROD1       INV01       2          05-01-2018
4    PROD1       INV03       1          05-01-2018
5    PROD2       INV02       3          08-01-2018
6    PROD2       INV04       4          08-01-2018
  1. Get daily summed amounts for each PRODx but fill the gaps with the previous ones if day does not exist.

This means:

 product_id    amount       date
   PROD1         2          01-01-2018
   PROD2         3          01-01-2018
   PROD1         2          02-01-2018
   PROD2         3          02-01-2018
   PROD1         2          03-01-2018
   PROD2         3          03-01-2018
   PROD1         2          04-01-2018
   PROD2         3          04-01-2018
   PROD1         3          05-01-2018
   PROD2         3          05-01-2018
   PROD1         3          06-01-2018
   PROD2         3          06-01-2018
   PROD1         3          07-01-2018
   PROD2         3          07-01-2018
   PROD1         3          08-01-2018
   PROD2         7          08-01-2018

A few thoughts:

  1. For first question I could obtain the max(date) for each PRODx and the pick for each PRODx the rows that have the date=with max(date) but I was wondering if there's faster way to obtain this given a large number of recors in the database

  2. For the second question, I could generate a series of dates for the interval needed and then use WITH rows As and do the query grouping by product_id and sum by amount and then select for each date the previous values from rows with a limit 1 but that does not sound that optimized either.

Looking forward for any input. Thank you.

Later edit: Trying to give DISTINCT ON () a try.

  • If I have distinct on(product_id, invoice_id) then I don't get only the most recent ones for the most recent date. If there were invoice_ids in the the past, beside the latest date, then they will be returned
  • If I have distinct on (product_id) then it returns from the most recent date, but as normal, only the last rows even if in the last day I have two positions for PROD1.

Basically I need something like 'I need for the most recent date, all the product_ids and their invoice_ids while keeping in mind that a product_id can have multiple invoice_ids'

Later edit 2:

Running a query like for first question seems to be reasonably fast:

select product_id, invoice_id, amount
from mytable inner join myOtherTable on...
             inner join (select max(date) as last_date, product_id 
                         from mytable 
                         group by product_id) sub on mytable.date = 
                         sub.last_date 
  • 1
    You can use DISTINCT ON for the first query: SELECT DISTINCT ON (product_id, invoice_id) id, product_id, invoice_id, amount, date FROM t ORDER BY product_id, invoice_id, date DESC ; – ypercubeᵀᴹ Nov 13 '17 at 22:10
  • Asked many times: dba.stackexchange.com/questions/tagged/… – a_horse_with_no_name Nov 13 '17 at 23:05
  • @ypercubeᵀᴹ I've edited the question with distinct on. Can you please take a look. Thank you. – Alin Nov 20 '17 at 14:01
  • @Alin it's not very clear why my suggestion does not fit your request. Do you want all the results from the latest date per product but all invoices (from that last date)? – ypercubeᵀᴹ Nov 20 '17 at 16:08
3

Skinning Q#1 independently and slightly differently than @ypercube

with cte as (select row_number() over (partition by product_id,
                                       invoice_id 
                                 order by dt desc) as rn,
                    product_id,
                    invoice_id,
                    amount,dt
               from product ) 
select product_id, invoice_id,amount,dt
  from cte
 where rn=1
 order by product_id,invoice_id;

 product_id | invoice_id | amount |     dt     
------------+------------+--------+------------
 PROD1      | INV01      |      2 | 2018-01-05
 PROD1      | INV03      |      1 | 2018-01-05
 PROD2      | INV02      |      3 | 2018-01-08
 PROD2      | INV04      |      4 | 2018-01-08
(4 rows)

For Q#2, you are on the right track, but the SQL will have a cross join(gasp!)

I think a function with a loop/cursor would be more optimized (i'll try that in my next free block of time)

--the cte will give us the real values
with cte as (select product_id, 
                    sum(amount) as amount, 
                    dt
               from product
              group by product_id,dt)
select p.product_id,  
       (select cte.amount --choose the amount
          from cte
         where cte.product_id = p.product_id
           and cte.dt <= d.gdt -- for same day or earlier
         order by cte.dt desc
         limit 1) as finamt,
       d.gdt
from (select generate_series( (select min(dt)
                                 from product), --where clause if some products 
                                                --don't have an amount
                              (select max(dt)
                                 from product),
                              '1 day' 
                            )::date as gdt)  d
cross join --assuming each listed product has an amount on the min date
     (select distinct product_id
        from product) p
left join --since we need to fill the gaps
     cte on ( d.gdt = cte.dt 
             and p.product_id = cte.product_id)
order by d.gdt, p.product_id
;
  • Awesome response, thanks. I've just implemented something similar with your cte example and suits my needs. Regards – Alin Dec 11 '17 at 13:50
2

I understand that you want all the rows with the latest date for every product (ties included, i.e. all rows with the last date). This can be done with rank() function:

select id, product_id, invoice_id, amount, date
from
  ( select id, product_id, invoice_id, amount, date,
           rank() over (partition by product_id
                        order by date desc) as rnk
    from 
        -- your joins
  ) as t 
where rnk = 1 ;
  • I've tried using rank() versus the query I've just added in question edit. While the query with rank is very readable, is about 5 times slower... – Alin Nov 27 '17 at 16:01
0

I agree your later edit way, it should be:

select product_id, invoice_id, amount 
    from mytable inner join 
    (select max(date) as last_date, product_id, invoice_id 
        from mytable 
        group by product_id) sub 
    on mytable.date = sub.last_date 
    and mytable.product_id = sub.product_id 
    and mytable.invoice_id = sub.invoice_id;

The "key" should be the date, product_id and invoice_id.

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