Return number of row in postgresql

Question

I have a where i store orderrows (salesrow_id), so one line of an invoice, which is part of a order order_id and the price paid for that item

salesrow_id(pk) | order_id | price | name
10                1           5      spaghetti
11                1           4      pizza
12                2           6      salami
13                1           1      vegetables

What i am trying to achieve is to have get the rownumber on the invoice, by order_id and salesrow_id. For example: salesrow_id=13, order_id=1 should return 3, since it's the third item on my invoice for order 1, if sorted by salesrow_id.

I've tried querying using row_number but that works only if i return multiple rows, and i would just like to have the number 3 as a single result.

Could someone push me in the right direction?

Gordon Linoff · Accepted Answer · 2019-01-09 16:30:32Z

1

You are looking for row_number(), I think:

select t.*,
       row_number() over (partition by order_id order by sales_row_id) as rownumber
from t;

If you want this for a single row, then use a subquery and filter:

select t.*
from (select t.*,
             row_number() over (partition by order_id order by sales_row_id) as rownumber
      from t
     ) t
where sales_row_id = 13;

If you care about performance, a correlated subquery might be better:

select t.*,
       (select count(*)
        from t t2
        where t2.order_id = t.order_id and t2.sales_row_id <= t.sales_row_id
       ) as rownum
from t
where t.sales_row_id = 13;

Note that row_number() (in the first query) is best for assigning results to the entire table. The correlated subquery would only have better performance when you are selecting one row or a smallish number of rows.

answered Jan 9, 2019 at 16:30

Gordon Linoff

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1 Comment

Marcovannoord Over a year ago

Ah, the keyword here is partition by. I have indeed looked at the row_number() extensively, but i couldn't find a way to do it per "group", which is exactly what the partition by does. You saved the day!

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