How can I return the number of count function?

Question

I need to test if my query returns 1 or 0 (so if there is a key matching the one entered)

Here's my code :

 $key=$_POST['key'];
 $queryKey = mysql_query("SELECT COUNT(*) FROM `smf_invites` WHERE `key` = '$key'");
 $query = mysql_num_rows($queryKey);

 if( !empty ($key))
 {
    echo 'You have entered a key';
    if (!empty ($query))
    {
        echo 'A key is corresponding';
    }

 }

EDIT : $connect = new mysqli("XXX","XXX","XXX","smf");

if (mysqli_connect_errno()) 
{
    printf("Connection failed : %s\n", mysqli_connect_error());
    exit();
}
else
{
    echo 'Connected to database';
}
$key=$_POST['key'];
if( !empty ($key))
{
 echo 'You have entered a key';
 $key = mysqli_real_escape_string($_POST['key']);
 $queryKey = mysqli_query("SELECT 1 FROM `smf_invites` WHERE `key` =   '$key'");
if (mysqli_num_rows($queryKey))
{
    echo 'A key is corresponding';
}
}
else
{
    echo 'No keys entered';
}

Now my code is giving me errors like

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in D:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\registration.php on line 32

Warning: mysqli_query() expects at least 2 parameters, 1 given in D:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\registration.php on line 33

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in D:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\registration.php on line 34

Answer 1

You have multiple issues with this code

1. If $_POST['key'] is not set you will get a PHP Notice
2. $queryKey = mysql_query("SELECT COUNT(*) FROM `smf_invites` WHERE `key` = '$key'"); will always return one row. So checking for the number of rows returned will not work.

Here is updated code that should resolve your issue.

 if( isset($_POST['key']))
 {
    echo 'You have entered a key';
    $key = mysql_real_escape_string($_POST['key']);
    $queryKey = mysql_query("SELECT 1 FROM `smf_invites` WHERE `key` = '$key'");
    if (mysql_num_rows($queryKey))
    {
        echo 'A key is corresponding';
    }
 }

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.

Your code is also wide open to SQL injections

Answer 2

In conjunction with John's answer

As per your newly edited code:

mysqli_ requires DB connection be passed for functions.

Change:

$key = mysqli_real_escape_string($_POST['key']);

to:

$key = mysqli_real_escape_string($connect, $_POST['key']);

or (use the variable $key in there instead.)

Then there's:

$queryKey = mysqli_query("SELECT 1 FROM `smf_invites` WHERE `key` =   '$key'");

to:

$queryKey = mysqli_query($connect, 
"SELECT 1 FROM `smf_invites` WHERE `key` =   '$key'") 
or die(mysqli_error($connect));

---

Troubleshooting/debugging

Add error reporting to the top of your file(s) which will help find errors.

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

Plus, or die(mysqli_error($connect))

Sidenote: Error reporting should only be done in staging, and never production.

Answer 3

I created a class for you and an example check this out.

<?php
class MyDb {
    protected $link;
    public function __construct() {
        if (!$this->link = mysql_connect('localhost', 'user', 'password')) {
            echo 'Could not connect to mysql';
            exit;
        }

        if (!mysql_select_db('table_name', $this->link)) {
            echo 'Could not select database';
            exit;
        }
    }

    public function exist($id) {
        if (is_int($id)) {
            $sql = "SELECT COUNT(*) FROM smf_invites WHERE `key` = '$id'";
            $result = mysql_query($sql, $this->link);
            return  mysql_num_rows($result);
        } else {
            return false;
        }
    }
}


$x = $_POST['key'] = 1;

if(!empty($_POST['key'])) {
    $db = new MyDb();
    if($db->exist($x)) {
        echo "exist";
    } else {
        echo "no exist";
    }
}