mysql data existence code not working [duplicate] - php

Question

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When to use single quotes, double quotes, and backticks in MySQL
(13 answers)
Closed 7 years ago.
i am trying to check a data existence from mysql table but following script not working. bellow my codes are provided please find out where is my mistake there.
<?php
//including the database files
include("../inc/settings.php");
$email = $_POST['email'];
$password = $_POST['password'];
$query = mysql_query("SELECT easy123 FROM users WHERE email=$email", $conn);
if (mysql_num_rows($query) != 0)
{
echo "Username already exists";
}
else
{
echo "this username not used";
}
?>
The error i am getting is-
Warning: mysql_query() expects parameter 2 to be resource, object
given in C:\xampp\htdocs\myfiles\Easy123\master\login.php on line 8
Warning: mysql_num_rows() expects parameter 1 to be resource, null
given in C:\xampp\htdocs\myfiles\Easy123\master\login.php on line 10
this username not used

Answer 1

First of all, make sure your database connection is correctly set up. The error you're getting clearly says that your $conn variable isn't a valid resource.
Also, use prepared statements and parameterized queries. Do not use PHP variables within your query string, it's not secure at all. Use instead PDO or MySQLi
Using PDO:
$stmt = $pdo->prepare('SELECT easy123 FROM users WHERE email = :email');
$stmt->execute(array('email' => $email));
foreach ($stmt as $row) {
// do something with $row
}
Using MySQLi:
$stmt = $dbConnection->prepare('SELECT easy123 FROM users WHERE email = ?');
$stmt->bind_param('s', $email);
$stmt->execute();
$result = $stmt->get_result();
while ($row = $result->fetch_assoc()) {
// do something with $row
}

Answer 2

Your $query seems to be wrong. Try this:
$query = mysql_query("SELECT easy123 FROM users WHERE email='$email'", $conn);
Make sure $conn is properly defined aswell.