如何显示重复的电子邮件地址 [英] how to display duplicate email address

问题描述

$sql ="SELECT u.Contact_Email FROM email1 AS u UNION ALL SELECT e.Contact_Email FROM email2 AS e"; $result = mysql_query($sql);
 $yes = 'yes';
 $no = 'no';
 echo "<table><tr><th>Email</th><th>email 1 opened</th><th>email 1 clicked</th><th>email2 opened</th><th>email2 clicked</th><th>email3 opened</th><th>email4 opened</th><th>email 4 clicked</th><th>email 5 opened</th><th>email 5 clicked</th><th>email 6 opened</th><th>email 6 clicked</th></tr><tbody>";
 while($row = mysql_fetch_array($result)) {
            $d_name = $row['Contact_Email'];
    ?> 
          <tr>
            <td><?php echo $d_name; ?></td>
            <?php if($row['Contact_Email']  == $row['Contact_Email']){
            echo "<td>yes</td>";
            } else if ($row['Contact_Email']  == 0){
            echo "<td>no</td>";
            }    ?>
            <td><?php echo $yes; ?></td>
          </tr>
          <?php } ?>
           </tbody>
    </table>

这是我的php脚本。我有一个11表格电子邮件1，电子邮件2，电子邮件3，...电子邮件11

连接表并显示重复的电子邮件为'是'否'否'
i想要这个输出我在做什么？ br>
像这样的输出

this is my php script. I have a 11 table email1, email2, email3, ... email11
join table and show duplicate email as 'yes' else 'no' i want this output whats i am doing??
Output like this

推荐答案

这是一个4表的示例，您可以将其扩展到11，如果需要的话。
对不起，我没有调试这个代码，我想主要的障碍是mysql查询从数据库中获取正确的值。

Here is an example for 4 tables you can extend it till 11 if you need. Sorry, but I did not debug this code, I guess the main obstacle was the mysql query to get correct values from database.

你绝对应该停止使用 mysql * 函数！

And you definitely should stop using mysql* functions!

$sql ="SELECT t.Contact_Email,
  e1.Contact_Email email1,
  e2.Contact_Email email2,
  e3.Contact_Email email3,
  e4.Contact_Email email4
FROM (
SELECT e.Contact_Email FROM email1 e
UNION ALL 
SELECT e.Contact_Email FROM email2 e
UNION ALL 
SELECT e.Contact_Email FROM email3 e
UNION ALL 
SELECT e.Contact_Email FROM email4 e
) t
LEFT JOIN email1 e1
ON t.Contact_Email = e1.Contact_Email
LEFT JOIN email2 e2
ON t.Contact_Email = e2.Contact_Email
LEFT JOIN email3 e3
ON t.Contact_Email = e3.Contact_Email
LEFT JOIN email4 e4
ON t.Contact_Email = e4.Contact_Email";

echo '<table><thead><tr><th>Email</th>';
for ($i=1;$i<5; $i++){
    echo "<th>email $i</th>";
}
echo '</tr></thead><tbody>';

$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
    echo '<tr><td>'.$row['Contact_Email'].'</td>';
    for ($i=1;$i<5; $i++){
        echo '<td>'.(empty($row['email'.$i])?'no':'yes').'</td>';
    }
    echo '</tr>';
}
echo '</tbody></table>';

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