On Thu, May 31, 2018 at 7:22 AM, Nicolas Seinlet <nico...@seinlet.com>
wrote:

> Hi,
>
> I have a query with a strange query plan.
>
> This query is roughly searching for sales, and convert them with a
> currency rate. As currency rate changes from time to time, table contains
> the currency, the company, the rate, the start date of availability of this
> rate and the end date of availability.
>
> The join is done using :
>     left join currency_rate cr on (cr.currency_id = pp.currency_id and
>           cr.company_id = s.company_id and
>           cr.date_start <= coalesce(s.date_order, now()) and
>          (cr.date_end is null or cr.date_end > coalesce(s.date_order,
> now())))
>
> The tricky part is the date range on the currency rate, which is not an
> equality.
>
> the query plan shows:
> ->  Sort  (cost=120.13..124.22 rows=1637 width=56) (actual
> time=14.300..72084.758 rows=308054684 loops=1)
>                           Sort Key: cr.currency_id, cr.company_id
>                           Sort Method: quicksort  Memory: 172kB
>                           ->  CTE Scan on currency_rate cr
> (cost=0.00..32.74 rows=1637 width=56) (actual time=1.403..13.610 rows=1576
> loops=1)
>
> There's 2 challenging things :
> - planner estimates 1637 rows, and get 300 million lines
> - sorting is generating lines
>

These are both explained by the same thing.  The sort is feeding into a
merge join.  For every row in the other node which have the same value of
the scan keys, the entire section of this sort with those same keys gets
scanned again.  The repeated scanning gets counted in the actual row count,
but isn't counted in the expected row count, or the actual row count of the
thing feeding into the sort (the CTE)


>
>
For now, the more currency rates, the slowest the query. There's not that
> much currency rates (1k in this case), as you can only have one rate per
> day per currency.
>

If it is only per currency per day, then why is company_id present? In any
case, you might be better off listing the rates per day, rather than as a
range, and then doing an equality join.

Cheers,

Jeff

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