How to find which emails are in the same lists?

前端 未结 2 453
野性不改
野性不改 2021-01-07 15:10

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



        
相关标签:
2条回答
  • 2021-01-07 15:19

    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.

    0 讨论(0)
  • 2021-01-07 15:33

    Since community 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>';
    
    0 讨论(0)
提交回复
热议问题