Question

我在PHP MYSQL上遇到了这个问题这是我的数据库样本

id word email
1  Helo jon@gmail.com
2  Sim  jon@gmail.com
3  Sam  jon@gmail.com
4  Mac  mars@gmail.com
5  Mimic mars@gmail.com

现在我试图在这里解决的是，我如何用相同的电子邮件计算电子邮件 OUTPUT网页：

id word email          submitted words dictionary
1  Helo jon@gmail.com        3         regular
2  Sim  jon@gmail.com        3         regular
3  Sam  jon@gmail.com        3         regular
4  Mac  mars@gmail.com       2         regular
5  Mimic mars@gmail.com      2         regular

提交的单词是电子邮件的计数。我怎样才能在php上计算出同样的电子邮件并输出3 ...

Answer 1

使用此查询：

SELECT email, COUNT(*) AS nb_emails
FROM your_table
GROUP BY `email`

它会为表格中的每封电子邮件返回相关记录的数量。

如果您使用的是mysql_* functions，则可以使用该代码获取所有值：

$sql = "SELECT email, COUNT(*) AS nb_emails
          FROM your_table
          GROUP BY `email`";
$res = mysql_query($sql);
while($row = mysql_fetch_assoc($res))
{
    // do something with the values of the last fetched record. The values are stored in $row['email'] and $row['nb_emails']
}

注意：不鼓励使用MySQL扩展。相反，应使用MySQLi或PDO_MySQL扩展名。有关详细信息，另请参阅MySQL: choosing an API指南和related FAQ。

Answer 2

使用mysql_num_rows()例如：

//mysql_connect() & mysql_select_db()    
$q=mysql_query("SELECT * FROM tablename WHERE email='jon@gmail.com'");
$count=mysql_num_rows($q); //3

Answer 3

试试这个：

SELECT a.ID, a.Word, a.Email, b.SubmittedWords
FROM   myTable a INNER JOIN
        (
            SELECT email, COUNT(*) SubmittedWords
            FROM mytable
            GROUP BY email
        ) b
            ON a.email = b.email

Answer 4

使用此：

select email,
count(email) as count_same
from table_users
group by email
having ( count(email) > 1 );

PHP Mysql检查行和有多少人有相同的电子邮件

4 个答案: