Re: BUG #17916: Expression IN list translates to unqualified operator
От | Tom Lane |
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Тема | Re: BUG #17916: Expression IN list translates to unqualified operator |
Дата | |
Msg-id | 2956974.1683121244@sss.pgh.pa.us обсуждение исходный текст |
Ответ на | BUG #17916: Expression IN list translates to unqualified operator (PG Bug reporting form <noreply@postgresql.org>) |
Ответы |
Re: BUG #17916: Expression IN list translates to unqualified operator
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Список | pgsql-bugs |
PG Bug reporting form <noreply@postgresql.org> writes: > create operator qwe.= (leftarg = char, rightarg = text, function = > qwe.chartexteq, commutator = operator(qwe.=), hashes, merges); > set search_path = qwe; > explain (costs off, verbose on) select i from generate_series(1, 10) i where > i::char in (2::text); > I expected, that IN list translates to pg_catalog.= Why would you expect that? It'd make it impossible to use IN with user-defined data types. In this case, you made an operator that is a closer match to the given datatypes (ie, "char = text") than the native "text = text" operator, so it used that one. I've not checked the code, but my recollection is that X IN (Y) just resolves to the same equality operator you'd get by writing X = Y. There's been some discussion about allowing a schema qualifier to be included in the syntax, but nothing's been done about that. regards, tom lane
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