Результат запроса SQL

258

06 августа 2017, 08:55

Я хочу что б мне вылезло результат ячейки которая совпадает с введенным мылом!

<html>
<body>
<center><form action='' method='POST'>
Email:<input type='text' name='email'><br/>
Manager:<input type='text' name='manager'><br/>
Date:<input type='text' name='date'><br/>
<input type='submit' value='CHECK'><br/>
</body>
</html>
<?php
    $hostmanager="****";
    $usermanager="****";
    $password="***";
    $dbmanager="****";
    $userstable="customers";
    $connect=@mysql_connect($hostmanager, $usermanager, $password) OR DIE("MySql Error"); mysql_select_db($dbmanager) or die(mysql_error());
    $name = (mysql_fetch_row(mysql_query("SELECT manager FROM customers WHERE email = '$email'")));
    if(isset($_POST['email'])){$email=$_POST['email'];}
    if(isset($_POST['manager'])){$manager=$_POST['manager'];}
    if(isset($_POST['date'])){$date=$_POST['date'];}
    if(empty($email) ||  empty($manager))
    {
        echo "Fill fields!";
    }
    if(@mysql_num_rows(mysql_query("SELECT email FROM $userstable WHERE email = '$email'"))==0){
        if(!empty($email) && !empty($manager)){
            if(isset($_POST['email'])){
                if($_POST['email']){
                    $sql=mysql_query("INSERT INTO $userstable (email, manager, date) VALUES ('$email', '$manager', '$date')");
                    $result=mysql_query($sql);
                    echo "Successfull atached to <b>$manager</b>";
                }       
            }
        }
    }
    else {echo "User already added by ".$name." ";}
    mysql_close();
?>

</center>
</html>

Answer 1

$email = mysql_real_escape_string($_POST['email']);
$result = mysql_query('SELECT `email`, `manager`, `date` FROM `customers` WHERE `email` = "' . $email . '"');
$data = mysql_fetch_assoc($result);
if (empty($data)) {
    die('email not found');
}
vprintf('
<pre>
email: %s
manager: %s
date: %s
</pre>', $data);