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How to count the number of records when using GROUP BY COUNT(*) DISTINCT_PHP tutorial

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Release: 2016-07-21 15:30:19
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For example, in a table like this, I want to count the number of records with different emails and passwords

Copy code The code is as follows:

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', ”);

Usually this is how we do it
Copy code The code is as follows:

SELECT COUNT(*) FROM test_users WHERE 1 = 1 GROUP BY email,passwords

What is the result of this?
Copy code The code is as follows:

COUNT(*)
1
2
1
1

Obviously this is not the result I want. What is counted is the sum of the number of records with the same email and passwords. The following is enough
Copy code The code is as follows:

SELECT COUNT(DISTINCT email,passwords) FROM `test_users` WHERE 1 = 1

Of course in php You can also use mysql_num_rows to get the number of records, but this is not efficient. You can refer to this article
mysql_num_rows VS COUNT efficiency problem analysis

www.bkjia.comtruehttp: //www.bkjia.com/PHPjc/323248.htmlTechArticleFor example, in a table like this, I want to count the number of records with different emails and passwords. Copy the code as follows: CREATE TABLE IF NOT EXISTS `test_users` ( ​​`email_id` int(11) unsigned...
source:php.cn
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