PHP include for mysql query without using my_sql_connect - php

Question

I'm attempting to create a dropdown list that is populated using PHP by gathering all of my clients in my database. This is working, but the problem is the code I am using requires the my_sql_connect and my_sql_select_db login details to populate the list.
On the pages this feature is required, the following snippet of code is provided at the top of each page to connect to the database:
require_once ('includes/dbConfig.php');
The code that is working is below (the #'s substitute my login details):
mysql_connect('localhost', '#', '#');
mysql_select_db('#');
$sql = "SELECT client_organisation_name FROM clients";
$result = mysql_query($sql);
echo "<select name='clients'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['client_organisation_name'] ."'>" . $row['client_organisation_name'] ."</option>";
}
echo "</select>";
What I am having difficulty in achieving is the dropdown list to populate using the connection already established. I have used the following code to try and substitute this but to no success:
$conn = mysql_connect('$dbhost', '$dbuser', '$dbpass') or die ('Error connecting to mysql');
mysql_select_db('$dbname');
I have searched the internet but have not found a suitable solution. For obvious reasons, it is a security risk and hassle to have your database login details evident, therefore I would be grateful for some help to get my snippet of code to connect to my database.

Answer 1

You can use MySQLi:
define(DB_HOST, "yourhost"); // Host
define(DB_NAME, "yourdbname"); // Database
define(DB_USER, "youruser"); // Username
define(DB_PASS, "yourpass"); // Password
$mysqli=new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME) or die("Error " . mysqli_error($link));
mysqli_set_charset($mysqli, 'utf8');
$results = $mysqli->query("SELECT client_organisation_name FROM clients");
while(($array_results[] = $results->fetch_assoc()) || array_pop($array_results));
echo '<select name="clients">'."\n";
foreach($array_results as $client) {
echo '<option value="'.$client['client_organisation_name'].'">'. $client['client_organisation_name'].'</option>'."\n";
}
echo '</select>'."\n";
Make sure that the connection data (user, password, name and host) are correct.

Answer 2

The single quote strings don't evaluate the variables inside.
So your code should be:
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');
mysql_select_db($dbname);
EDIT:
I saw #Jason McCreary's comment and I have to agree with him.
You should definitely use mysqli. The only changes in your code are following:
$db = new mysqli($dbhost, $dbuser, $dbpassword, $dbname);
Then you execute the mysql queries like this:
$db->query("INSERT MYSQL CODE HERE");