php - How to display my output in a dropdown list -

i want program list data mysql table in list in dropdown menu on page. here's code:

   <fieldset> <legend> selecteer uw categorie </legend>  <label   ="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 connect mysql: " . mysqli_connect_error();   }  $result = mysqli_query($con,"select * 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, looks data need , posts in dropdown list gets posted in single line.. want listed underneath each other.. please me!

they on single line because put result in 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 connect mysql: " . mysqli_connect_error()."</option>";         }          $result = mysqli_query($con,"select * categorien");          while($row = mysqli_fetch_array($result))         {              echo "<option>".$row['categorieen1'] . "</option>";         }          mysqli_close($con);     ?>   </select> </datalist>  </fieldset> 

edit:

sorry mistake forgot second echo in while loop xp!