python - psycopg2:使用外键插入时出错

Question

我试图首先在表“lf_calculator_user”上插入一行，然后获取我刚刚插入的行的 ID，并使用该 ID 将其作为外键插入到表“lf_calculator_request”的另一条记录中。 这是我的代码:

 if event["email"] is not None:
        cursor.execute("INSERT INTO lf_calculator_user (email) VALUES (%(email)s) ON CONFLICT (email) DO UPDATE SET email = EXCLUDED.email;", event)
        cursor.execute('SELECT LASTVAL()')
        lastid = cursor.fetchone()[0]
        if lastid is not None: 
            if event["amount"] is not None: 
                if event["date"] is not None:
                    cursor.execute('INSERT INTO lf_calculator_request (email_id, amount, invoice_due_date) VALUES ('+ str(lastid) +', %(amount)s, %(date)s);', event)

我收到一个奇怪的错误，说 email_id 不是表“lf_calculator_user”上的列，显然不是，它是作为外键的“lf_calculator_request”上的列。我错过了什么？

这里是错误:

psycopg2.IntegrityError: insert or update on table "lf_calculator_request" violates foreign key constraint "lf_calculatorrequest_email_id_fkey"DETAIL:  Key (email_id)=(21) is not present in table "lf_calculator_user".

Answer 1

在 Postgresql 中尝试更详细的日志记录，问题可能会变得更加明显。 更新 postgresql.conf(并重启)

log_min_duration_statement = -1 # -1 禁用，0 记录所有语句

log_min_duration_statement = 0