Checking if email is in Mysql database in PHP

Question

I am trying to figure out how to check if an email is already within a database as I don't want duplicates I have tried googling and it hasn't worked as old posts I think?

here's my code:

$emailcheck = $value2;
$emailcheck = "SELECT email FROM demo WHERE email= '$emailcheck' ";
$result = mysql_query($link, $emailcheck);
$count = mysql_num_rows($result);

if ($count > 0){
    {$_SESSION['email'] = "Email is already used!";}    
}



if (empty($_POST["name"]))
     {$_SESSION['name']= "Name is required";}
   if (empty($_POST["email"]))
     {$_SESSION['email'] = "Email is required";} else {
   if (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL))
       {$_SESSION['email'] = "Email is incorect";}}



if(empty($_POST["name"]) || empty($_POST["email"]) || (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)) || ($count > 0)) 
{   
    header("Location: home.php");
} else {

thank you for anyones help!

My nearly full code:

$value = $_POST['name'];
$value2 = $_POST['email'];

$emailcheck = $value2;
$emailcheck = "SELECT email FROM demo WHERE email= '$emailcheck' ";
$result = mysql_query($link, $emailcheck);
$count = mysql_num_rows($result);

if ($count > 0){
    {$_SESSION['email'] = "Email is already used!";}    
}



if (empty($_POST["name"]))
     {$_SESSION['name']= "Name is required";}
   if (empty($_POST["email"]))
     {$_SESSION['email'] = "Email is required";} else {
   if (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL))
       {$_SESSION['email'] = "Email is incorect";}}



if(empty($_POST["name"]) || empty($_POST["email"]) || (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)) || ($count > 0)) 
{   
    header("Location: home.php");
} else {


if (!$link) {
    die('Could not connect: ' . mysql_errno());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) {
    die('can\'t use ' . DB_NAME . ':' . mysql_error());
}

$sql = "INSERT INTO demo (name, email) VALUES ('$value', '$value2')";

if (!mysql_query($sql)) {
    die('Error: ' . mysql_error());
}
header("location: https://www.google.co.uk/search?q=Ghostbusters&stick=H4sIAAAAAAAAAONgFuLUz9U3MI43skxR4gIxjUxLcrLTtYQck0vyi0LyQ8Ic81J888syU4sBdwGq3i0AAAA&sa=X&ved=0ahUKEwj0-J6QtuXaAhUQSsAKHaYqB2wQxA0I6gEwHg&biw=1280&bih=699");     
mysql_close();
session_destroy();
}
}
?>

whats the question? your piece of code is not even closing right :( — Tin, May 03 '18 at 19:52
in my code i my form is still letting me go though the process if there is a duplicate email within the database — Adam Martin, May 03 '18 at 19:53
What is `$value2`? How can you store 2 different values in the same variable, `$emailcheck`? What error(s) are you getting? — NewBee, May 03 '18 at 19:54
it not giving me errors it just passing me though to the website using the header for the sucessful entry and value2 is the name of the email from the from so the email ? — Adam Martin, May 03 '18 at 19:56
You need to stop/quit the process in this condition true `if ($count > 0){ {$_SESSION['email'] = "Email is already used!";} }` Right now even if it's a duplication email, it continues the process. — Tin, May 03 '18 at 19:56
Also two things: currently your SQL code is unsafe and allows SQL injections, please take a look to [How can I prevent SQL injection in PHP?](https://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) and consider to use `PDO` or `mysqli` to avoid it, and remember that `mysql_*` extension is obsolete, and you should change it to (again) `PDO` or `mysqli` instead. — Julian David, May 03 '18 at 20:34

score 0 · Answer 1 · answered May 03 '18 at 19:56

0

I think you have to put

die;

after

header("Location: home.php");

otherwise the code will still continue executing

answered May 03 '18 at 19:56

Peshraw H. Ahmed

439
3
22

can you just change header("Location: home.php"); to header("Location: home.php");die; and tell me if it works? You are on the right way but you are not stoping the execution after redirecting to home. – Peshraw H. Ahmed May 03 '18 at 20:09

Malovich · Answer 2 · 2018-05-03T20:17:41.757

Peshaw and Tin are absolutely correct; there's nothing to alter or stop code execution even if you discover a duplicate email.

You will know if there is a duplicate, but your code won't respond to that as is currently written.

You'll need something after you store the result in session...

$isDuplicate = false;

if ($count > 0){
  $_SESSION['email'] = "Email is already used!";
  $isDuplicate = true;
}

if(empty($_POST["name"]) || empty($_POST["email"]) || (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)) || ($count > 0) || ($isDuplicate)) {
  // error handling here
  header("Location: home.php");
} else {
  // regular code execution here
  if (empty($_POST["name"]))
    {$_SESSION['name']= "Name is required";}
  if (empty($_POST["email"]))
    {$_SESSION['email'] = "Email is required";} else {
  if (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL))
    {$_SESSION['email'] = "Email is incorrect";}}
}

...and of course this answer won't be complete without the usual warning of sticking user input straight into queries instead of parameterizing them with PDO or at least sanitizing them. Please sanitize your inputs.

i dont know if im adding it right please may you try add it into my code im sorry? — Adam Martin, May 03 '18 at 20:09
i have added this and i just seems to be carrying on when it shouldnt any ideas? — Adam Martin, May 03 '18 at 20:26
is there anyway of me sending you both my scripts and you testing it on your end? — Adam Martin, May 03 '18 at 20:27

score 0 · Answer 3 · answered May 03 '18 at 20:04

0

You need to wrap the rest of your code after if ($count > 0) in an else code block, like this:

if ($count > 0){
    $_SESSION['email'] = "Email is already used!";

} else {
    if (empty($_POST["name"]))
        {$_SESSION['name']= "Name is required";}
        // the rest of your code
}

answered May 03 '18 at 20:04

Julian David

311
1
12

1

This will work too. It depends on how involved Adam's process is. – Malovich May 03 '18 at 20:05

Checking if email is in Mysql database in PHP

3 Answers3