PHP - Making Options Available Change Based On What Is Selected From Dropdown Box

Hi freaks,

 I'm new to php first of all.  I'm dynamically binding a dropdownlist with mysql database . After the user selects an item from it , I want to match that item with another table so as to populate another database.

The code I'm using to populate dropdown:
Code: [Select]
<?php
 $con = mysql_connect("localhost","root","");
if(!$con)
{
  die ('Can not connect to : '.mysql_error());
}
mysql_select_db("ims",$con);
$result=mysql_query("select cat_id,cat_name from category");
echo "<select name=cat>";
while($nt=mysql_fetch_array($result))
{
 echo "<option value=$nt[cat_id]> $nt[cat_name] </option>";
}
echo "</select>";
        mysql_close($con);
  ?>

Now after the user selects any one of the item , I want to bind another dropdown on the same page using such query like
   $result=mysql_query("select subcategory.sc_id,subactegory.sc_name from subcategory,category where subcategory.sc_id=$nt[cat_id]");

Please anyone tell me the logic and code to do it. Also tell me do I need an intermediate page to  post the 1st dropdown value and then continue with 2nd dropdown. I couldn't figure out the concept anyhow.
Help on this will be highly appreiable .
(Tell me if I'm not clear with my question)

I'm trying to figure out why the options aren't appearing inside the select dropdown. Any ideas why?

Code: [Select]
echo "<label for=" . $row2['fullName'] . ">" . $row2['fullName'] . "</label>";
                       echo "<select name=" . $row2['fullName'] . " id=" . $row2['fullName'] . " class=dropdown title=" . $row2['fullName'] . " />";
                       if ($styleID == 1 || $styleID == 2 || $styleID == 6) {
                         $charactersQuery = "
                                SELECT
                                    characters.ID,
                                    characters.characterName
                                FROM
                                    characters
                                WHERE
                                    characters.styleID = 3
                                ORDER BY
                                    characters.characterName";
                         $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query
                         while ( $row3 = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) {
                            print "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r";
                         }   
                       } else {
                         $charactersQuery = "
                                SELECT
                                    characters.ID,
                                    characters.characterName
                                FROM
                                    characters
                                WHERE
                                    characters.styleID IN (1,2,6)
                                ORDER BY
                                    characters.characterName";
                         $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query
                         while ( $row3 = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) {
                            print "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r";
                         }
                       }
                       echo "</select>";
                   }

I understand that is might be something that is already answered and I apologize if it is, I could not find it. What I need to do is build a simple form that has two options, they will be dropdown options. Dropdown A and Dropdown B then a Submit button. This part I understand in HTML, although it may be easier in php or javascript. Then I need it to take the two options and create a "if/then" statement that loads a specific pdf that matches the two options selected. Example. If someone selects Option 1 from Dropdown A and Option 2 From Dropdown B then it loads 12.pdf
If someone selects Option 5 from Dropdown A and Option 3 From Dropdown B then it loads 53.pdf
If someone selects Option 2 from Dropdown A and Option 1 From Dropdown B then it loads 21.pdf and so on... It does not have to be the exact thing just some way to take both inputs and have it equal a specific pdf. Here is the form I built but I don't know what to put in the form_action.php file in order to make it work

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<center>
<h1> Get Directions</h1>
<form action="form_action.php" method="get" name="directions" target="_new">

<select name="startpoint" size="1">
<option value="north">North Tower Entrance</option>
  <option value="south">South Tower Entrance</option>
  <option value="moba">MOB A Entrance</option></select>
  
  ----->

<select name="endpoint" size="1">
<option value="onco">Oncology</option>
  <option value="radio">Radiology</option>
  <option value="pulm">Pulmanary</option></select>
<br /><br />
<input type="submit" value="Submit" />
</form>
</center>

</body>
</html>

Any help is appreciated, thanks.

Hi im new to php and I need help making webpage that queries a mysql database based on a 3 check boxes and displays results on the same page or on another page. The table being queried has 4 columns, name, gps, wifi, bluetooth. So for example a row in the table would be like, samsung galaxy s2, yes, yes, yes. The idea is for it to be a website that will display phones according to their features.

So the idea is depending on if the boxes were ticked the samsung galaxy would be displayed as a result. So i need some help understanding how to make this.

Some1 gave me the code below in attempt to help me (im not sure it works or not) but im not sure how fully use it, ie what pages i need to make and how i create the connection to the mysql database, and how to use the query that they wrote to display the results

thanks

code:
Code: [Select]
<form action="?do=filter" method="post">
<table cellspacing="0" cellpadding="3" border="1">
<tr>
<td>GPS<input type="checkbox" name="gps" value="checked"></td>
<td>Wifi<input type="checkbox" name="wifi" value="checked"></td>
<td>Bluetooth<input type="checkbox" name="bluetooth" value="checked"></td>
</tr>
<tr><td><input type='submit' name='filter' value='Filter'></td></tr>
</table>
</form>
</html>
<?php
function filterMe($filter){
if(isset($_POST[$filter])){
return "Yes";
}else{
return "No";
}
}

if(isset($_POST['filter'])){
echo "Gps - " . filterMe('gps');
echo "
Wifi - " . filterMe('wifi');
echo "
Bluetooth - " . filterMe('bluetooth');
}
?>


All you need to do is use a query something like
SELECT name,gps,wifi,bluetooth FROM `product` WHERE `gps`='".filterMe('gps')."' AND `wifi`='".filterMe('wifi')."' AND `bluetooth`='".filterMe('bluetooth')."'

Hi

I have been trying to get a value to be selected in a mysql populated dropdown list but can't get it to work and was hoping someone could help

I have a database with user info in it and this is an update page where they can update their details. The code i have (which doesn't work) is:

<select name="agency">
<? $query1 = mysql_query("SELECT * FROM agents ORDER BY agent ASC",$connect);
while($myrow = mysql_fetch_assoc($query1)){
$agent = $myrow['agent'];
echo "<option"; if ($agent == $agency) { echo "selected='selected'"; } echo ">$agent</option>";
}
?>
</select>

The $agency value is the current agency which is stored in the users profile and the value does exist in the list which is being populated (also, i have define $agency further up in my code)  so i don't know why the selected value won't display. No value is displayed in the dropdown list on the page - but the values are in the list if i remove the selected='selected' part of the code.

Any help yould be greatly appreciated.

Merry Christmas
Andy