How to find which emails are in the same lists?

问题:

I have a 11 tables [email1, email2, email3, ... email11]

<?php
$con = mysql_connect("localhost", "root", "");
$db = mysql_select_db("email-db", $con);
$sql = "SELECT Contact_Email FROM email1, email2, email3, email4, email5";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
    ?>
    <tr>
        <td><? echo $row['Contact_Email']; ?></td>
        <td><? echo '<br>'; ?></td>
    </tr>
<? } ?>

What I actually want to do it to select all emails from all tables using join on all of them. How can it be done?

回答1:

MySQL UNION operator allows you to combine two or more result sets from multiple tables into a single result set

SELECT Contact_Email FROM email1
UNION
SELECT Contact_Email FROM email2
UNION
SELECT Contact_Email FROM email3
.
.
UNION
SELECT Contact_Email FROM email11

By default, the UNION operator eliminates duplicate rows from the result even if you don’t use DISTINCT operator explicitly.

回答2:

$sql = "SELECT Contact_Email FROM email1 UNION ALL SELECT Contact_Email FROM email2 UNION ALL SELECT Contact_Email FROM email3 UNION ALL SELECT Contact_Email FROM email4 UNION ALL SELECT Contact_Email FROM email5 UNION ALL SELECT Contact_Email FROM email6 UNION ALL SELECT Contact_Email FROM email7 UNION ALL SELECT Contact_Email FROM email8 UNION ALL SELECT Contact_Email FROM email9 UNION ALL SELECT Contact_Email FROM email10 UNION ALL SELECT Contact_Email FROM email11";  

i used this

回答3:

Since society voted to close duplicate how to display duplicate email address

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.

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>';

UPDATE Use the same query but change the output part to:

echo '<table><thead><tr><th>Email</th>';
for ($i=1;$i<5; $i++){
    echo "<th>email $i</th>";
}
echo "<th>Total (yes)</th>";
echo "<th>Total (no)</th>";
echo '</tr></thead><tbody>';

$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
    $yesCount = 0;
    $noCount = 0;
    echo '<tr><td>'.$row['Contact_Email'].'</td>';
    for ($i=1;$i<5; $i++){
        echo '<td>'.(empty($row['email'.$i])?'no':'yes').'</td>';
        if (empty($row['email'.$i])) {
            $noCount++;
        } else {
            $yesCount++;
        }
    }
    echo '<th>'.$yesCount.'</th>';
    echo '<th>'.$noCount.'</th>';
    echo '</tr>';
}
echo '</tbody></table>';