BUG #16400: IN (query) allows for reference to column that doesn't exist
От | PG Bug reporting form |
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Тема | BUG #16400: IN (query) allows for reference to column that doesn't exist |
Дата | |
Msg-id | 16400-84f68bb46ba963e5@postgresql.org обсуждение исходный текст |
Ответы |
Re: BUG #16400: IN (query) allows for reference to column thatdoesn't exist
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Список | pgsql-bugs |
The following bug has been logged on the website: Bug reference: 16400 Logged by: Scott Marcy Email address: postgresql@mscott.org PostgreSQL version: 12.2 Operating system: Mac OS Description: The following code should generate an error for an unknown column in the SELECT query that uses IN (), but it does not, and it behaves as though all rows pass the IN test. A standalone SELECT on the missing column generates an error as it should. I've attached two examples, one using a view, one using a temporary table, both of which exhibit the same problem. Note, however, that using another column name like "foo" does properly generate an error at the IN subquery. -- Sample using view BEGIN; CREATE TABLE test_table ( id SERIAL PRIMARY KEY, first_name text, last_name text, email text ); INSERT INTO test_table (first_name, last_name, email) VALUES ('Alice', 'Able', 'a.able@example.com'), ('Bob', 'Bubble', 'b.bubble@example.com'); SELECT COUNT(*) FROM test_table; -- 2 CREATE VIEW test_view AS SELECT id AS user_id, email FROM test_table WHERE first_name LIKE 'A%'; SELECT COUNT(*) FROM test_view; -- 1 SELECT COUNT(*) FROM test_table WHERE id IN (SELECT id FROM test_view); -- Should be error, "id" does not exist in test_view, but outputs 2. SELECT id FROM test_view; -- Proof "id" does not exist in test_view ROLLBACK; -- Sample using temporary table BEGIN; CREATE TABLE test_table ( id SERIAL PRIMARY KEY, first_name text, last_name text, email text ); INSERT INTO test_table (first_name, last_name, email) VALUES ('Alice', 'Able', 'a.able@example.com'), ('Bob', 'Bubble', 'b.bubble@example.com'); SELECT COUNT(*) FROM test_table; -- 2 CREATE TEMPORARY TABLE temp_test_table AS SELECT id AS user_id, email FROM test_table WHERE first_name LIKE 'A%'; SELECT COUNT(*) FROM temp_test_table; -- 1 SELECT COUNT(*) FROM test_table WHERE id IN (SELECT id FROM temp_test_table); -- Should be error, "id" does not exist in temp_test_table, but outputs 2. SELECT id FROM temp_test_table; -- Proof "id" does not exist in temp_test_table, will generate error as expected ROLLBACK;
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