通過連接提高MySQL嵌套選擇性能

Question

我已經看到了許多示例，這些示例可通過聯接來改善MySQL嵌套選擇，但對於此查詢，我無法弄清楚：

SELECT * FROM messages WHERE answer = 'SuccessSubscribed' AND phone NOT IN 
         (SELECT phone FROM messages WHERE answer = 'SuccessUnSubscribed');

該查詢查找已訂閱但從未取消訂閱的人員。

表結構：

CREATE TABLE `messages` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`phone` varchar(12) COLLATE utf8_persian_ci NOT NULL,
`content` varchar(300) COLLATE utf8_persian_ci NOT NULL,
`flags` int(10) unsigned NOT NULL DEFAULT '0',
`answer` varchar(50) COLLATE utf8_persian_ci DEFAULT NULL,
....,
PRIMARY KEY (`id`),
....
) ENGINE=InnoDB CHARSET=utf8 COLLATE=utf8_persian_ci

Answer 1

可以使用帶有NULL檢查的LEFT JOIN來代替NOT IN 。

SELECT M1.*
FROM messages M1
LEFT JOIN messages M2 ON M2.phone = M1.phone AND M2.answer = 'SuccessUnSubscribed'
WHERE M1.answer = 'SuccessSubscribed' AND M2.phone IS NULL