PHP check database for user ID

Question

I'm sending data to a php doc with jquery ajax. The first check ($check_user_exists) on the user's inputs is to see if there's an account under the same $email. I've tried writing a function before the if statement but that didn't work either.

I'm getting the following error:

Parse error: syntax error, unexpected 'return' (T_RETURN) in C:\xampp\htdocs\workflow\ajax\register.php on line 8

Thanks ahead of time! Here's the code:

if (isset($_POST['email']) && isset($_POST['firstname']) && isset($_POST['lastname']) && isset($_POST['role']) && isset($_POST['pw'])){
    $email = strtolower($_POST['email']);
    $firstname = $_POST['firstname'];
    $lastname = $_POST['lastname'];
    $pw = crypt($_POST['pw'], md5($email));
    $role = $_POST['role'];
    $check_user_exists = return (mysql_result(mysql_query("SELECT `user_id` FROM `users` WHERE user_id = '$email.'"), 0)==1) ? true : false;
    if($check_user_exists === true){
        echo 'Our records show an account already exists under this email.';
    }

Using mysql_num_rows() would cut down your code by no less than 30%. — Funk Forty Niner
– Funk Forty Niner, Commented Jul 31, 2014 at 23:59
"Didn't work" is not good enough. Explain what the problem is. — Sverri M. Olsen
– Sverri M. Olsen, Commented Aug 1, 2014 at 0:00
mysql_result() returns the contents of the user_id field. Since your user IDs are apparently their email, not an identifier number, it will never be == 1. — Barmar
– Barmar, Commented Aug 1, 2014 at 0:00

Barmar · Accepted Answer · 2014-08-01 00:25:20Z

1

Do it like this:

$email = mysql_real_escape_string(strtolower($_POST['email']));
$firstname = mysql_real_escape_string($_POST['firstname']);
$lastname = mysql_real_escape_string($_POST['lastname']);
$pw = crypt($_POST['pw'], md5($email));
$role = mysql_real_escape_string($_POST['role']);
$result = mysql_query("SELECT COUNT(*) AS count FROM `users` WHERE user_id = '$email'") or die(mysql_error());
$row = mysql_fetch_assoc($result);
$check_user_exists = $row['count'];
if ($check_user_exists > 0){
    echo 'Our records show an account already exists under this email.';
} else {
    mysql_query("INSERT INTO users (user_id, firstname, lastname, pw, role)
                 VALUES ('$email', '$firstname', '$lastname', '$pw', '$role')") or die(mysql_error());
    echo 'User added successfully.';
}

edited Aug 1, 2014 at 0:25

answered Aug 1, 2014 at 0:07

Barmar

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2 Comments

user2937995 Over a year ago

Hey Barmar, looks like it works in the sense that it doesn't add any new entries but it doesn't echo the error message. I forgot to mention that my database is called 'workflow' and it's hosted on localhost. so I added a select db statement, but it still doesn't work

Barmar Over a year ago

You had an extra . after $email in the query. I've added the code to add the new entry. One of the two messages should be echoed.

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