php global array help

[blizbiggy]

I cant seem to get my $dropdown_value to display. I have tried lots of things to make the array global or even try to call functions, but it is difficult when you are calling 2 tables within an array while looping... Here is the original code... It displays ONLY ONE value of the "admins" and has the correct number of dropdown options from the database, but it does not give values or display names for the last them.

<?php
include("../include/config.php");

$checkout_connect = mysql_connect($dbhost,$dbUsername,$dbPassword);
$selectdb1 = mysql_select_db($dbDatabase,$checkout_connect);
if (!$checkout_connect || !$selectdb1){
echo "Could <b>NOT</b> connect to database. Please check the config.php!";
die;
}
/*Begin Query*/
if(!isset($_POST['submit']))
{
?>
<form name="form" action="<?php echo $PHP_SELF;?>" method="post">
  <table width="700" border="0">
    <tr>
      <td width="200"><div align="right">Item Name:&nbsp;</div></td>
      <td width="100"><input type="text" name="name"><font size=1> Example: Colorjet Ink</font></td>
      <td width="400" align="left"></td>
    <tr>
      <td><width="200"><div align="right">&nbsp;</div></td>
      <td><input type="submit" value="submit" name="submit" /></td>
      <td width="400" align="left"></td>
    </tr>

  </table>
</form>
<?php
}
$count=0;

//Do this once you click submit
if(isset($_POST['submit']))
{

//Begin Dropdown Option Grab
$sql_dropdown = "SELECT * FROM checkout_user WHERE admin='1'";
$sql_dropdown_result = mysql_query($sql_dropdown) or die('Query failed: ' . mysql_error());
$numrows = mysql_num_rows($sql_dropdown_result);
if($numrows == 0)
{
        echo "None Found";
}
else
{
        for($i=0; $i<$numrows; $i++)
        {
                while ($dropdown_search = mysql_fetch_array($sql_dropdown_result))
                {
                        $dropdown_value = array($i => array(dropdown_name => $dropdown_search['name'],
                                                         dropdown_barcode => $dropdown_search['barcode']));
                }
        }
}


//Begin Second Query and Display values
$sql = "SELECT * FROM checkout_item WHERE itemname like \"%$_POST[name]%\" OR barcode like \"%$_POST[name]%\"";
$result = mysql_query($sql) or die('Query failed: ' . mysql_error());

while ($search = mysql_fetch_array($result)){
$value = array($count => array(itemname => $search['itemname'],
                             barcode    => $search['barcode'],
                             returncond => $search['return_condition'],
                             added      => $search['date_checked'],
                             notes      => $search['notes'],
                             status     => $search['is_it_checked_out'],
                             checkedout => $search['checked_out_by'],
                             checkedin  => $search['checked_in_by'],
                             userchk    => $search['checked_out_to'],
                             used       => $search['times_used'])
              );
echo $value[$count]["itemname"];


//Display Dropdown Options
echo "<select name='select'>";
for($i=0; $i<$numrows; $i++)
{
        $OPTION=$dropdown_value[$i]["dropdown_name"];
        $VALUE=$dropdown_value[$i]["dropdown_barcode"];
        echo "<option value=$VALUE>$OPTION</option>";
}
echo "</select><br>";
$count++;
}
}
?>

[blizbiggy]

I figured out my problem... and simplified my code a bit... I was told you can do two mysql_fetch_arrays within each other because they use different resources. So, I went back through and tried to figure out why mine only displayed once. Here is my final code:

<?php
include("../include/config.php");

$checkout_connect = mysql_connect($dbhost,$dbUsername,$dbPassword);
$selectdb1 = mysql_select_db($dbDatabase,$checkout_connect);
if (!$checkout_connect || !$selectdb1){
echo "Could <b>NOT</b> connect to database. Please check the config.php!";
die;
}

/*Begin Query*/
if(!isset($_POST['submit']))
{
?>
<form name="form" action="<?php echo $PHP_SELF;?>" method="post">
  <table width="700" border="0">
    <tr>
      <td width="200"><div align="right">Item Name:&nbsp;</div></td>
      <td width="100"><input type="text" name="name"><font size=1> Example: Colorjet Ink</font></td>
      <td width="400" align="left"></td>
    <tr>
      <td><width="200"><div align="right">&nbsp;</div></td>
      <td><input type="submit" value="submit" name="submit" /></td>
      <td width="400" align="left"></td>
    </tr>

  </table>
</form>
<?php
}
if(isset($_POST['submit']))
{

$sql = "SELECT * FROM checkout_item WHERE itemname like \"%$_POST[name]%\" OR barcode like \"%$_POST[name]%\"";
$result = mysql_query($sql) or die('Query failed: ' . mysql_error());

$count=0;
while ($search = mysql_fetch_array($result)){

echo $search['itemname'];
echo " ";

echo "<select name='select'>";

$sql_dropdown = "SELECT * FROM checkout_user WHERE admin='1'";
$sql_dropdown_result = mysql_query($sql_dropdown) or die('Query failed: ' . mysql_error());
$numrows = mysql_num_rows($sql_dropdown_result);
if($numrows == 0)
{
        echo "None Found";
}
else 
{
        $count2 = 0;
        while ($search2 = mysql_fetch_array($sql_dropdown_result))
        {
                $OPTION=$search2['name'];
                $VALUE=$search2['barcode'];
                echo "<option value=$VALUE>$OPTION</option>";
        $count2++;
        }
}
echo "</select><br>";
$count++;
}
}
?>