Populating a dropdown list through SQL with specific condions - php

Question

Essentially, i'm trying to select all the users from my table where their type ='users' and if that condition is met, select their first and last name. Store them into variables and then display the variables into a dropdown list one by one(array).
Current code:
<div id="ContainerC">
<form action="AppointmentAddQuery.php" method="post">
<input type="text" name="date" class="datepicker" placeholder="Please select a date" required><br>
<input type="text" name="patient" class="textbox" placeholder="Patient" required> <br>
<?php
include 'dbconnection.php';
?>
<input type="text" name="phlebotomist" class="textbox" placeholder="phlebotomist" required> <br>
<input type="text" name="bloods" class="textbox" placeholder="blood required" required><br>
<p>Has the appointment been allocated?</p>
<select name="allocated" class="textbox" required>
<option value="yes">yes</option>
<option value="no">no</option>
</select>
<br><br><br>
<button type="submit" name="submit" class="SUB">Add patient</button>
</form>
</div>
SQL setup:

Answer 1

For the SQL part of your question, you are probably looking for something like:
select firstname
, lastname
from <table>
where type = 'user'

Answer 2

Assuming the structure of your database (you will likely need to replace table and column names) this is how you could do it:
<?php
include 'dbconnection.php';
$sql = "SELECT firstname, lastname FROM users WHERE type = 'users'";
$result = $conn->query($sql) or die($conn->error);
if ($result->num_rows > 0) {
echo '<select>';
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<option value="'.$row["firstname"].$row["lastname"].'">'.$row["firstname"].$row["lastname"].'</option>';
}
echo '</select>';
} else {
echo '<select><option value="">No Users</option></select>';
}
$conn->close();
?>
This also assumes that inside of dbconnection.php you created a mysqli object like so:
$conn = new mysqli($servername, $username, $password, $dbname);