display select option from database when select option change

Question

<tr>
  <td><label class="label label-success">Building Name: </label></td>
  <td>
    <select class="form-control" name="building_name" required>
      <option value="">Select Building</option>
      <?php
      include '../Database/db.php';
      $res=mysql_query("SELECT * FROM building");
      while($row = mysql_fetch_array($res)){
      ?>
      <option value="<?php echo $row['building_name']; ?>"><?php echo $row['building_name']; ?></option>

      <?php } ?>
    </select>
  </td>
</tr>
<tr>
  <td><label class="label label-success">Floor: </label></td>
  <td>
    <select class="form-control" name="floor" required>
      <option value="">Select Floor</option>
      <option>></option>

    </select>
  </td>
</tr>

can anyone help me i want to display the floor of the selected building from the database. if i choose the building the floor will be display on select option floor.

where the floors info is stored? db structure will be helpful as well. — mitkosoft, Feb 26 '16 at 10:04
it is like when i choose country the province will be on a selected option — jayson, Feb 26 '16 at 10:07

score 1 · Answer 1 · answered Feb 26 '16 at 10:07

Without knowing your database setup or structure here's is an example of how you could achieve this:

<?php require_once('../Database/db.php'); ?>
<tr>
  <td><label class="label label-success">Building Name: </label></td>
  <td>
    <select class="form-control" name="building_name" required">
      <option value="">Select Building</option>
      <?php
      $res=mysql_query("SELECT * FROM building");
      while($row = mysql_fetch_array($res)){
      ?>
      <option value="<?php echo $row['building_name']; ?>"><?php echo $row['building_name']; ?></option>
      <?php } ?>
    </select>
  </td>
</tr>
<tr>
  <td><label class="label label-success">Floor: </label></td>
  <td>
    <select class="form-control" name="floor" required>
      <option value="">Select Floor</option>
      <?php
      $building_name = mysql_real_escape_string($_POST['building_name']);
      $res2=mysql_query("SELECT * FROM floor WHERE building_id = '{$building_name}'");
      while($row = mysql_fetch_array($res2)){
      ?>
      <option value="<?php echo $row['floor_name']; ?>"><?php echo $row['floor_name']; ?></option>
      <?php } ?>
    </select>
  </td>
</tr>

You need to pass the posted parameter to an WHERE clause in the SQL. Note that you should be escaping malicious code form the posted parameters.

I'm also assuming that you have this html code inside a form tag along with a submit button

just curious with this approach - how you will execute second PHP part of code (floor related) without refreshing the page after selecting the buiding above? — mitkosoft, Feb 26 '16 at 10:09
@mitkosoft He is not asking how to achieve this without refreshing the page. — karlingen, Feb 26 '16 at 10:10

Gopalakrishnan · Accepted Answer · 2016-02-26T10:53:32.773

0

You should use AJAX to get the database value to display in DROPDOWN

$('#building_name').change(function(){
var building_name=  $('#building_name').val();
$.ajax({
url: "index.php", //Change it yours
type: "post",
data: {building_name: building_name}, 
success: function(data) {
    if(data != "false")
{
    $('#Floor').html(data); 
}
else{
alert("No value");}
}
});
});

And PHP FUNCTION should be

  $building_name = mysql_real_escape_string($_POST['building_name']);
  $res2=mysql_query("SELECT * FROM building  WHERE building_name = '{$building_name}'");
  while($row = mysql_fetch_array($res2)){
  echo '<option value="'.$row['floor_name'].'">'.$row['floor_name'].'</option>';
  }

edited Feb 26 '16 at 10:53

answered Feb 26 '16 at 10:07

Gopalakrishnan

957
8
19

Yes when you select – Gopalakrishnan Feb 26 '16 at 10:25
What you have done. Because I have given Sample Code for recruitment. Change it yours like table name, column name change it – Gopalakrishnan Feb 26 '16 at 10:34
this are my columns name, building_name, floor, location and the table name is building – jayson Feb 26 '16 at 10:39
the floor is on the table building – jayson Feb 26 '16 at 10:43
OK you want floor drop down display according to building drop down option? or you want display all floor in building table? – Gopalakrishnan Feb 26 '16 at 10:44
I have edit the php function in above function. Try it – Gopalakrishnan Feb 26 '16 at 10:54
i copy the codes but it didn't work. it think the problem is on ajax – jayson Feb 26 '16 at 11:02
Did use $(document).ready(function() { }); in start on javascript ? – Gopalakrishnan Feb 26 '16 at 11:03
nothing. i select of select option building but when i click floor there is no data appear. – jayson Feb 26 '16 at 11:05
press F12 after select option building. There would display a error. Find it and tell me – Gopalakrishnan Feb 26 '16 at 11:06
Did you change the url in ajax as it yours? – Gopalakrishnan Feb 26 '16 at 11:09
did you add id in – Gopalakrishnan Feb 26 '16 at 11:10
$('#floor').html(data); i think it is the problem. – jayson Feb 27 '16 at 14:20

display select option from database when select option change

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