Anyone who can assist me with my assignment

Question

its a web database where we are asked to populate drop-down tables retrive data from phpMyAdmin must be able to edit n delete mine just don't add data in the tables using the add form

Answer 1

Create a user on your SQL server which only has these Data permissions:
"SELECT (for reading)
UPDATE (for editing)
DELETE"
Create a database for it too. (remember the username and password, don't use root)

Create your database PHP config file with your database details along with your new username and password you created.
Should look something like:
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT id, data";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Data: " . $row["firstname"]. " " . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>

Then you can instead of echoing the data, rather put them in your dropdown list using HTML and a while loop.
PHP+HTML
while ($row = $sql->fetch_assoc()){
echo "<option value=\"data\">" . $row['data'] . "</option>";

<option value="Data"><?php echo $row['data']; ?></option>

and you can use this html+php combo file i found:
<div id="contact_form">
<form action="contact_insert.php" method="post">
Firstname: <input type="text" name="firstname">
Lastname: <input type="text" name="surname">
Email: <input type="text" name="email">
<input type="submit" name="submit">
</form>
</div>

<?php
if(isset($_POST['submit']))
{
$username="root";
$password="password";
$server="127.0.0.1";
$database="eddiesdb";

$con = mysql_connect($server,$username,$password);

$sql="INSERT INTO customer (FirstName, Surname, EmailAddress)
VALUES
('$_POST[firstname]','$_POST[surname]','$_POST[email]')";
$a=mysql_query($sql);?>

Hope this solution can help you.