Forms, dynamic drop down list

[Rulkster]

If someone can shed some light on this or point me in the right direction, i would be very grateful. Here is my dilema:

i have two tables:

MAIN CATEGORY
Category id, name

1, glass
2, stone
3, iron

SUB CATEGORY
SubCategoryid, name, categoryid

1, vase, 1
2, vase, 2
3, candleholder,2
4,candleholder,1

What i want in my form is two drop down lists, i can populate the first one with the Main Category Names, so when the user selects say Glass, then the second drop down list will be populated with all of the sub categories for glass (in this case: vase, candleholder).

This should all take place before the user clicks the submit button. I've seen some java scripts that can do this but you have to hardcode the selections and i need to get them from a table where i don't know what they are. Is there anything in PHP that can handle this.

Thanks for any help

[planetsim]

As PHP is server side everything is done on the server. So unless a form is being submitted nothing will happen for PHP to do this.

Like you said. Javascript can do it. As its client side, it reacts with what the client's(visitor's) browser. So if you want something to populate. So in Javascript. Use PHP code in it. Its the only way i can think of it working. Until you want just Javascript Doing it. Meaning Updating/deleting/adding categories wont work.

[jowatkins]

OK, you need to do a few things here:

first, lets populate your first drop down box from the database. Put this dropdown box within a form of it's own, then we can submit the information to be used in the second drop down.

<form action="page.php" name="form1" action="post">

<?
$query1_result = mysql_query ("SELECT catagory FROM table");
$num_results = mysql_num_rows($query1_result );
$catagories = array();
//use the for loop to populate the array with the unitnos from the query
for ($i=0; $i<$num_results; $i++)
{
  $catagory_row=mysql_fetch_object($query1_result );
  $catagories[]=$catagory_row->catagory;
}

//start the dropdown box
echo "<select name=\"catagory\" value=\"$catagory\" onChange=\"document.form1.submit();\">"; 
//detect if this script has been submited yet
if (!isset ($catagory)) 
{
  $selected_catagory = "";
}
else
{
  $selected_catagory = $catagory;
}
if ($selected_catagory == "")
{
  echo "<option value=\"\" SELECTED>SELECT CATAGORY</option>";
}
else
{
  echo "<option value=\"\" >SELECT CATAGORY</option>";
}
//populate the option and provide a SELECTED option for each item in the array

foreach ($catagories as $catagory)
{	
  if ($selected_catagory == $catagory) 
  {
    echo "<option value=\"$catagory\" SELECTED >$catagory</option>";
  }
  else
  {
    echo "<option value=\"$catagory\" >$catagory</option>";
  }
}
echo "</select>";
?>

</form>

this script will select all the catagories in your db, and populate the options in your drop down box with the results. It also detects if the catagory value has been submitted yet, and if so, mantains that value in the dropdown list.

All you need to do now is write an if statement to see if you have a set catagory value and repeat the process for building the second drop down box, altering your SELECT statement to include a WHERE clause for the catagory id. If you mean to submit this information on to a further page, then you will need to put a hidden field into your second form to hold the $catagory value.

Hope this helps, Jo