Another small problem: It seems with PostgreSQL you must always qualify
your datasource name with a schema name, otherwise you will get a
confusing IndexError. In order to get a better error message, I suggest
replacing the following code in dbPostgreSQL
tableNameBreak=tableName.split(".", 1)
localSchemaName = tableNameBreak[0]
localTableName = tableNameBreak[1]
with this (appears 2 times):
try:
localSchemaName, localTableName = tableName.split(".", 1)
except ValueError:
raise ValueError("Please use schema-qualified datasource names"
" (e.g. 'public.%s')" % tableName)
Another option would be to determine the schema name automatically using
the current_schemas() and is_visible() functions of Postgres, or to
check for the first matching name in getTables(). But that would mean
the application will be dependent on the search path (which may be good
or bad). Another option would be to simply assume "public", but "In the
face of ambiguity, refuse the temptation to guess."
-- Christoph
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