Update mysql query with reading HTML SELECT - php

Question

I'm making an orderform, just an internal one.
My form can already add a new order and summarize it on the internal page.
Form is made in a table.
Database queries (which are all displayed in the table): ordernr, klantnaam (customername), productnaam (productname), productid, status
Now, I'd like to change the status (with the HTML select option) by.. lets say "Not ordered" to "Ordered". and when I press the submit button it should "update" it in the SQL database.
HTML select list: "Niet besteld, Besteld, Onderweg naar hoofdlocatie, Onderweg naar vestiging, Ontvangen".
How do I do this?
$con=mysqli_connect("localhost","root","","bestelformulier");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM overzicht");
//table heading
echo "<table align='center' width='700px' border='2'>
<tr>
<th width='10px'>Ordernr</th>
<th width='10px'>Klantnaam</th>
<th width='10px'>Productnaam</th>
<th width='10px'>ProductID</th>
<th width='10px'>Status</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['ordernr'] . "</td>";
echo "<td>" . $row['klantnaam'] . "</td>";
echo "<td>" . $row['productnaam'] . "</td>";
echo "<td>" . $row['productid'] . "</td>";
echo "<td><select><option>" . $row['status'] . "</option></select></td>";
//end table
mysqli_close($con);
//adding order table
echo "<h5>Bestelling Toevoegen</h5>";
echo "<form method='post'>";
echo "<table width='700px' border='1'>
<tr>
<th>klantnaam</th>
<td><input type='text' name='klantnaam'/></td>
</tr>
<tr>
<th>Productnaam</th>
<td><input type='text' name='productnaam'/></td>
</tr>
<tr>
<th>productid</th>
<td><input type='text' name='productid'/></td>
</tr>
<tr>
<th>Status</th>
<td>
<select name='status'>
<option value='Niet besteld'>Niet besteld</option>
<option value='Besteld'>Besteld</option>
<option value='Onderweg naar hoofdlocatie'>Onderweg naar hoofdlocatie</option>
<option value='Onderweg naar vestiging'>Onderweg naar vestiging</option>
<option value='Ontvangen'>Ontvangen</option>
<select>
</td>
</tr>
<tr>
<td></td>
<td><input type='submit' name='submit' value='Toevoegen'/></td>
</tr>";
echo "</table>";
echo "</form>";
$klantnaam = $_POST['klantnaam'];
$productnaam = $_POST['productnaam'];
$productid = $_POST['productid'];
$status= $_POST['status'];
if(isset($_POST['submit'])) {
$query = mysqli_query($con,"INSERT INTO overzicht (klantnaam, productnaam, productid, status)
VALUES ('$klantnaam', '$productnaam', '$productid', '$status')");
$current_url = (empty($_SERVER['HTTPS']) ? "http://" : "https://") . $_SERVER['HTTP_HOST'] . $_SERVER['REQUEST_URI'];
header ('Location: ' . $current_url);
exit ();
}
mysqli_close($con);
This is the code I have been using so far.
Thanks in advance!

Answer 1

to create a select with all options
change this
echo "<td><select><option>" . $row['status'] . "</option></select></td>";
to
echo "<td><select>
<option>" . $row['status'] . "</option>";
if($row['status'] != "Niet besteld")
echo "<option>Niet besteld</option>";
if($row['status'] != "Besteld")
echo "<option>Besteld</option>";
if($row['status'] != "Onderweg naar hoofdlocatie")
echo "<option>Onderweg naar hoofdlocatie</option>";
if($row['status'] != "Onderweg naar vestiging")
echo "<option>Onderweg naar vestiging</option>";
if($row['status'] != "Ontvangen")
echo "<option>Ontvangen</option>";
echo" </select></td>";