Обсуждение: Bug in SQLForeignKeys()

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Bug in SQLForeignKeys()

От
Constantin S. Svintsoff
Дата:
Query used for checking foreign key triggers
returns too many results when there're more than one foreign
key in a table. It happens because only table's oid is used to
link between pg_trigger with INSERT check and pg_trigger with
UPDATE/DELETE check.

I think there should be enough to add following conditions
into WHERE clause of that query:
    AND    pt.tgconstrname = pg_trigger.tgconstrname
    AND    pt.tgconstrname = pg_trigger_1.tgconstrname

/Constantin
--- cut here ---

/fjoe