Solved

populate 2nd drop down based on 1st drop down

Posted on 2009-03-31
3
909 Views
Last Modified: 2013-12-12
Hi Thankyou both for your prompt response

I appoligise I should of explained there is only one mysql table called 'Resale' I cant change this and use a separate country table.

Here are the fields in the table I need to use:

Country , City

So I need to select the Country drop down which then populates the second city drop down automatically.

heres what I have got so far


<?php

$conn stuff here

$query = "SELECT DISTINCT Country FROM Resale ";
$result = mysql_query($query) or die ( ' error in query ' );


then populate the 1st drop down box

then need to refresh the page to populate the second city drop own box

$query = "SELECT DISTINCT City FROM Resale WHERE Country = (Country from 1st drop down)

please help

thanks
barry

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Question by:bazconnolly
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3 Comments
 
LVL 14

Accepted Solution

by:
shobinsun earned 500 total points
ID: 24027264
Hello,

Build some arrays in javascript that list the possible options, then fill the second dropdown with one of those arrays based on the selection in the first dropdown.

For more details, go through the links (an example):

http://www.dhtmlgoodies.com/index.html?whichScript=ajax_chained_select

http://www.dhtmlgoodies.com/scripts/ajax-chained-select/ajax-chained-select.html



http://teamtutorials.com/web-development-tutorials/pulling-drop-down-values-from-a-database-using-javascript-and-php

http://www.zazzybob.com/onchange.html

http://www.javascriptkit.com/jsref/select.shtml

Hope these links will help you.

Regards.
0
 
LVL 4

Expert Comment

by:Fugas
ID: 24027370
Better to use Ajax call to populate the second dropdown. Here is the code. So just download prototype.js from http://www.prototypejs.org/download
city.php
<?php
  $c=mysql_connect('localhost','root','');
  mysql_select_db('devel');
  $query = "SELECT DISTINCT City FROM Resale WHERE Country = '".$_GET['country']."'";
  $result = mysql_query($query) or die ( ' error in query ' );
  while($row = mysql_fetch_row($result)) echo '<option>'.$row[0].'</option>';
?>
index.php
<script src="prototype.js" type="text/javascript"></script>
<?php
  $c=mysql_connect('localhost','root','');
  mysql_select_db('devel');
  $query = "SELECT DISTINCT Country FROM Resale ";
  $result = mysql_query($query) or die ( ' error in query ' );?>
  <select onchange="new Ajax.Updater('city','http://ajax/city.php?country=' + this.options[this.selectedIndex].value)">;<?php
  while($row = mysql_fetch_row($result)) echo '<option>'.$row[0].'</option>';
  echo '</select>';
  echo '<select id="city"></select>';
?>
  

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0
 

Author Closing Comment

by:bazconnolly
ID: 31564743
excllent service thankyou
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