Re: BUG #18652: Planner can not find pathkey item to sort for query with expression and expression index - Mailing list pgsql-bugs

From Alena Rybakina
Subject Re: BUG #18652: Planner can not find pathkey item to sort for query with expression and expression index
Date
Msg-id d7ff8f5a-e64c-4bc9-a902-9795d3645df1@postgrespro.ru
Whole thread Raw
In response to Re: BUG #18652: Planner can not find pathkey item to sort for query with expression and expression index  (Andrei Lepikhov <lepihov@gmail.com>)
List pgsql-bugs
On 10.10.2024 13:23, Andrei Lepikhov wrote:
> On 10/10/24 16:02, Tender Wang wrote:
>>
>>
>> Andrei Lepikhov <lepihov@gmail.com <mailto:lepihov@gmail.com>> 于2024年 
>>      > I think this is because find_computable_ec_member() only tries to
>>      > match Vars.  Maybe we should teach it to also match OpExprs?
>>     Looking into that case, I don't understand only one thing:
>>     generate_orderedappend_paths decided to try MergeAppend; the
>>     create_append_path routine added the Sort cost,  but the Sort node
>>     itself wasn't added. Maybe the origin problem is the lack of
>>     feasibility
>>     examinations?
>>
>>
>> Yeah, I'm also curious why only cost Sort but not adding Sort node in 
>> create_merge_append_path().
>> The comments say that " We'll need to insert a Sort node, so include 
>> cost for that".  Does another place
>> insert the Sort node?
> Before inserting the Sort node, we must identify the column 
> corresponding to each path key. And here is the problem: IndexScan has 
> two elements in the target list, but SeqScan has only one. It already 
> looks strange to me. How do we UNION two sources with different 
> numbers of resulting columns? It seems to me we have a bug under the 
> Append node.
>
Maybe a mistake exists at the stage of creating a sequential scan? 
Because the subpath of the MergeAppend path does not have any elements 
in pathkeys, although the pasthtarget has one element and this 
expression wasn't involved anywhere?

I honestly don’t understand at what stage this expression i+1 appears:

SELECT i + 0 AS c FROM (SELECT i FROM t UNION ALL SELECT i + 1 FROM t)

-- 
Regards,
Alena Rybakina
Postgres Professional




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