Populating Lists

[ppaul]

I am using sql to connect to a database and then php to populate list from it, as you can see the second list is populated the decision of the first, now what Im trying to do is when a user selects from the second list I would like to fire a php statement which puts the result in one table -

<p>please select a store.</p>
  <select name="menu" id="myStore">
   <?php 
    include("scripts/dbconnect.php"); 
    $query="select distinct store from game order by store asc";
    $result=mysql_query($query);
    $numrows=mysql_num_rows($result);
    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
     echo '<option value="'.$row['store'].'">'.$row['store'].'</option>';
    }  

   ?>
   </select><input type="button" onclick="getStore()" value="Submit" /><br/>

   <?php
    if(isset($_GET['myStore'])){
     echo '<p>Select a game.</p>';
     echo '<select name="menu" id="myGame">';
     include("scripts/dbconnect.php"); 
     $myStore=$_GET['myStore'];
     $query="select description from game where store='$myStore'";
     $result=mysql_query($query);
     $numrows=mysql_num_rows($result);
     while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
       echo '<option value="'.$row['description'].'">'.$row['description'].'</option>';
     } 
     echo '</select>';
     echo '<input type="button" table()" value="Submit" /><br/>';
    }
   ?>

THIS IS WORKING ALL FINE ^{^^^^^^^^^^^^}

so based on the decision of the 2nd list I would like to fire this statement below and put the answer in a table -

   <?php
    if(isset($_GET['myGame'])){
     include("scripts/dbconnect.php"); 
     $myGame=$_GET['myGame'];
     $query="select id from game where description='$myGame'";
     $result=mysql_query($query);

print "<table border='1'>";
while ($a = mysql_fetch_array($res)){
  print "<tr>
        <td>$a[0]</td>
    </tr>";  }
print "</table>";


}
   ?>

I know this is a lot to take in, I appreciate any help, thank you

[Derokorian]

I wonder if

      echo '<input type="button" table()" value="Submit" /><br/>'; 

is causing the problem. You seem to have a stray table()" maybe you meant to do onchange="gettable()" or something similar?

Also this script is very vulnerable to SQL injection. More on that -> http://us3.php.net/manual/en/security.database.sql-injection.php

You also never check to verify that the query returns any results before trying to use it. See the link in my signature about mysql errors to see how you should be checking for this.

[bradgrafelman]

Since you're talking about client-side behavior, you're going to have to use a client-side scripting language. So... either submit the form when the first dropdown selection is made, or use something like AJAX to make a background request to a PHP script that receives input and returns options to be used in the second dropdown.

For the latter, try searching for phrases like "AJAX chained dropdown" or "AJAX chained select" or something of that nature.