How to display only result in page - php

Question

I have a big problem. I wrote a simple $_GET system
$query = mysql_query("SELECT * FROM `users`");
while ($row = mysql_fetch_array($query)) {
if($_GET['user'] == $row['user_seo']) {
echo 'user exists';
} else {
echo 'No users found';
}
}
If user don't exists in table it display 'No users found' otherwise it display No users found user exists No users found. 2 times it display "No users found" but user exists with that seo in database table. Thanks.
database screen http://prntscr.com/2ddqu4

Answer 1

You are looping over all your users so you will get multiple messages.
You should add a WHERE condition to your query to check only for the required user and switch to PDO or mysqli with prepared statements.
Something like (in PDO):
$query = 'SELECT * FROM `users` WHERE `user_seo`=:user';
$db->prepare($query);
$db->execute(array(':user' => $_GET['user']));
// etc.

Answer 2

Try replacing mysql_fetch_array with mysql_fetch_assoc
Also I would strongly recommend using either PDO or mysqli and prepared statements

Answer 3

Safely find the user with mysqli:
if (isset($_GET['user'])) {
$user = $_GET['user'];
$connection = mysqli_connect($Host, $Username, $Password) or die(mysqli_error());
mysqli_select_db($connection, $database) or die(mysqli_error());
$user = mysqli_real_escape_string($connection, $user);
if (!is_numeric($user)) {
$user = "'$user'";
}
$sql = "SELECT * FROM `users` WHERE `user_seo`=$user;";
$result = mysqli_query($connection, $sql);
if ($result) {
$user = mysql_fetch_assoc($result)) {
if ($user) {
// User found
// Do something with info like:
$userName = $user['name'];
} else {
// User NOT found
}
mysql_free_result($result);
} else {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
}

Answer 4

You better may use an sql like this:
SELECT * FROM users WHERE user_seo='".$_GET['user']."'; And forget about looping all the rows.
EDIT: but better use PDO or Mysqli and prepared statements