例如這樣一個表,我想統計email和passwords都不相同的記錄的條數spa
CREATE TABLE IF NOT EXISTS `test_users` ( `email_id` int(11) unsigned NOT NULL auto_increment, `email` char(100) NOT NULL, `passwords` char(64) NOT NULL, PRIMARY KEY (`email_id`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=6 ; INSERT INTO `test_users` (`email_id`, `email`, `passwords`) VALUES (1, ‘jims@gmail.com', ‘1e48c4420b7073bc11916c6c1de226bb'), (2, ‘jims@yahoo.com.cn', ‘5294cef9f1bf1858ce9d7fdb62240546′), (3, ‘default@gmail.com', ‘5294cef9f1bf1858ce9d7fdb62240546′), (4, ‘jims@gmail.com', 」), (5, ‘jims@gmail.com', 」);
一般咱們的作法是這樣:code
SELECT COUNT(*) FROM test_users WHERE 1 = 1 GROUP BY email,passwords
這樣的結果是什麼呢?blog
COUNT(*) 1 2 1 1
顯然這不是我要的結果,這樣統計出來的是相同email和passwords的各個記錄數量之和,下面這樣就能夠了:rem
SELECT COUNT(DISTINCT email,passwords) FROM `test_users` WHERE 1 = 1