Gregory Stark skrev:
> Nis Jørgensen <nis@superlativ.dk> writes:
>
>> Well, the query can be satisfied by looking only at the rows with an
>> order_id matching the invoice_id given. The condition that this is the
>> largest invoice in the group then needs to be checked afterwards.
>>
>> I certainly did not expect the query planner to be able to deduce this,
>> though.
>
> No, that's not true. If you had two records in eg_order with the same order_id
> but different invoice_ids then the query would need both records to satisfy
> the query.
I assume you mean "... then both records are necessary in order to
calculate the results of the query". This does not contradict what I wrote.
If you mean "... then both records need to satisfy <some criteria>" then
I don't understand which criteria you are talking about. The query in
question was:
SELECT order_id,max(order_view.invoice_id),sum(order_view.mileage) FROM (SELECT order_id,invoice_id, 0 as
mileageFROM eg_order UNION SELECT order_id, 0, mileage FROM eg_order_line) order_view
GROUPBY order_view.order_id;
This is then restricted on max(invoice_id)
As far as I can tell, these steps produce the correct results (without
the later information about primary keys provided by Bryce)
INPUT: my_invoice_id
1. Look up all order_ids for which (order_id,my_invoice_id) appear in
eg_orders
2. Find all rows (in both branches of the UNION) with these id_s
3. Group the rows, and calculate max(invoice_id)
4. Filter the result rows on max(invoice_id) = my_invoice_id.
> The query planner can't deduce that this can't happen because it simply does
> not have that information.
Well, I realize that the existing query planner can't. Since I could
arrive at the plan above by deduction so could a hypothetical different
query planner. Whether it is faster is of course unknown - my guess is
that it would be in this case.
> The more I look at this view the more I think it's just seriously broken.
> Why is it grouping by order_id at all if, I suspect, there will only be one
> record per order_id in eg_orders??
Bryce has confirmed this. The above is only of academic interest.
Yours,
Nis Jorgensen