Início / dba / Perguntas / 76642
Accepted
Gili
Asked: 2014-09-16 19:46:35 +0800 CST

Aviso do Postgresql: tipo de referência X convertido em Y (Estado SQL: 00000 - Código de erro: 0)

  • 772

Quando eu executo:

CREATE VIEW foreign_keys AS
SELECT
    tc.table_name, kcu.column_name, ccu.table_name AS foreign_table_name,
    ccu.column_name AS foreign_column_name
FROM
    information_schema.table_constraints AS tc, information_schema.key_column_usage AS kcu,
    information_schema.constraint_column_usage AS ccu
WHERE
    tc.constraint_type = 'FOREIGN KEY' AND tc.constraint_name = kcu.constraint_name AND
    ccu.constraint_name = tc.constraint_name;

CREATE FUNCTION GetPermissionColumns() RETURNS TABLE(column_name foreign_keys.column_name%TYPE)
    AS $$
BEGIN
    RETURN QUERY SELECT column_name FROM foreign_keys WHERE table_name = tg_table_name AND
        foreign_table_name = 'permission';
END;
$$ LANGUAGE plpgsql;

no Postgresql 9.3, recebo a seguinte mensagem de aviso:

type reference foreign_keys.column_name%TYPE converted to information_schema.sql_identifier (SQL State: 00000 - Error Code: 0)

O que estou fazendo de errado e como posso corrigir?

postgresql postgresql-9.3

1 respostas

  1. Best Answer

    Respondendo a minha própria pergunta...

    Acontece que este é um registro inofensivo no nível da INFORMAÇÃO: https://stackoverflow.com/a/3531274/14731

    • 0

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