How to display my output in a dropdown list - php

Question

I want my program to list my data from mysql table in a list in a dropdown menu on my page.
Here's my code:
<fieldset>
<legend> Selecteer uw Categorie </legend>
<label for ="Categorie"> Categorie </label>
<select name ="Categorie" id="Categorie">
<datalist id ="Categorie">
<Option Value="Router">Router</option>
<Option Value="Switch">Switch</option>
<Option Value="Toestel">Toestel</option>
<Option Value="Basisstation">Basisstation</option>
<Option Value="Repeaters">Repeaters</option>
<Option Value= <?php
$con=mysqli_connect("localhost","root","admin","inventarisdb");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Categorien");
while($row = mysqli_fetch_array($result))
{
echo "<ul>";
echo "<li>" . $row['Categorieen1'] . "</li>";
echo "</ul>";
}
echo "</table>";
mysqli_close($con);
?>
</option>
</select>
</datalist>
</fieldset>
THis code works perfectly, It looks up the data i need and posts it in the dropdown list But it all gets posted in a single line..
I want it to be Listed underneath each other..
Please help me!

Answer 1

They are all on a single line because you put your result in a single <option> tag.
try this:
<Option Value="Basisstation">Basisstation</option>
<Option Value="Repeaters">Repeaters</option>
<?php
$con=mysqli_connect("localhost","root","admin","inventarisdb");
// Check connection
if (mysqli_connect_errno())
{
echo "<option>Failed to connect to MySQL: " . mysqli_connect_error()."</option>";
}
$result = mysqli_query($con,"SELECT * FROM Categorien");
while($row = mysqli_fetch_array($result))
{
echo "<option>".$row['Categorieen1'] . "</option>";
}
mysqli_close($con);
?>
</select>
</datalist>
</fieldset>
EDIT:
sorry for the mistake I forgot the second echo in the while loop XP!