MySQL和PHP的SQL語法錯誤。 '選項從poll_maker_options WHERE poll_id = 3'

Question

我有這個疑問：

SELECT votes, option 
    FROM poll_maker_options 
WHERE poll_id = " . $row['poll_id'] 

我看沒有錯，但我得到的錯誤：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'option FROM poll_maker_options WHERE poll_id = 3' at line 1

感謝您的任何幫助。我希望我不會像昨天那樣失明，因爲我找不到一個錯字。

Answer 1

「選項」是MySQL中的reserved word，所以要麼更改字段名稱，或用反引號括起來在所有查詢它用於：

SELECT votes, `option` FROM ...