PHP MYSQLi query only accesses user on first row - php

Question

Follow the links above, in sequence, for an example of a username & password login web based system, developed using PHP, JavaScript and a MySQL database.
Follow the links above, in sequence, for an example of a username & password login web based system, developed using PHP, JavaScript and a MySQL database.
Follow the links above, in sequence, for an example of a username & password login web based system, developed using PHP, JavaScript and a MySQL database.

Answer 1

You are testing only the first row of the table user. When you test using the if statement, the first row returned is either the user trying to log in, or not.
Instead, you can do the following:
function user_exists($username) {
$username = sanitize($username);
$query = mysqli_query($_POST['x'], "SELECT * FROM users where username ='".$username."'");
$row = mysqli_fetch_array($query);
if(count($row)>0) {
return true;
}
else return false;
}
This looks for all rows with the given username. Of course, testing passwords or other things will depend on actually iterating through the array fetched, instead of just checking the count.

Answer 2

I finally did it yesterday thanks to you guys. It works well however is needed to be repeated in other functions involved (which is what failed) and yes I forgot my error reporting was turned off, making life harder for myself:
function user_exists($username) {
$username = sanitize($username);
$query = mysqli_query($_POST['x'], "SELECT `username` FROM `users` WHERE `username` = '$username'");
while($row = mysqli_fetch_assoc($query)){
$rows[] = $row;
}
if (count($rows) > 0) {
return true;} else return false;
}
Same done for the login function:
function login($username, $password){
$username = sanitize($username);
$password = md5($password);
$query = mysqli_query($_POST['x'], "SELECT `userid` FROM `users` WHERE `username` = '$username' AND `password` = '$password'");
while($row = mysqli_fetch_array($query)){
$rows[] = $row;
}
if(count($rows) > 0) {
$userid = user_id_from_username($username);
return $userid; } else return false;
Thanks again!