Selecting e-mail address from database

Question

I have this code selecting a combined e-mail and password but when I recall the e-mail using a login form the code fails and give me this error:

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]:
Syntax error or access violation: 1064 You have an error in your SQL syntax;
check the manual that corresponds to your MySQL server version for the right syntax to use near '@gmail.com = 'admin'' at line 1' in C:\wamp\www\PWS\index.php on line 15
( ! ) PDOException: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax;
check the manual that corresponds to your MySQL server version for the right syntax to use near '@gmail.com = 'admin'' at line 1 in C:\wamp\www\PWS\index.php on line 15'

I know why this isn't working though it recognizes the E-mail as a value Thodor20@gmail.com where @gmail.com is seperated because of the @. So how can I change the existing code:

<?php
    include("connect.php");        
    $logemail = @$_POST['email'];
    $logww = @$_POST['wachtwoord'];        
    if(isset($_POST['submit'])){
        $q2 = $db->prepare("SELECT * FROM userinfo WHERE $logemail = '$logww'");
        $q2->execute(array(':email'=>$logemail,':wachtwoord'=>$logww));
        echo "Login succesvol!";
    }
?>

And the form:

<form method="post">
    E-mail:<input type="text" name="email"><br>
    Wachtwoord:<input type="password" name="wachtwoord"><br>
    <input type="submit" name="submit" value="Inloggen"><br>
</form>

So it will accept Thodor20@gmail.com as 1 full value and not in parts?

Answer 1

$q2 = $db->prepare("SELECT * FROM userinfo WHERE logemail = :email");

When using prepared statements you can use question marks or named placeholders ie :email

Also you may want to add another where to select where logemail = :email and Wachtwoord = : Wachtwoord

Answer 2

I think you are using wordpress. I dont know exact syntax for wordpress query. But here you have syntax error in your select query.

Remove $ from $logemail after WHERE clause -

$q2 = $db->prepare("SELECT * FROM userinfo WHERE logemail = '$logemail'");