PHP Question

Ok well i've created a php script that basically echo's all of the tables within a database into a select box so that a user can select the table to use it then sets the $table variable to the selection that they have made once they click submit.

here is the code i used.

$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);

if (!$result) {
    echo "DB Error, could not list tables\n";
    echo 'MySQL Error: ' . mysql_error();
    exit;
}
echo "<form method=\"post\" action=\"" . $_SERVER['PHP_SELF'] ."\">"; 
echo "<span class=\"style\">Please Select a Category:</span> &nbsp;&nbsp;&nbsp;<select name=\"dbtables[]\" id=\"dbtables\">\n";
echo "<option value=\"\"></option>";
$accept = array();
while ($row = mysql_fetch_row($result)) {

    echo "<option value=\"".$row[0]."\">".$row[0]."</option>\n";
    $accept[$row[0]] = true;
}

if(!isset($accept[$_POST['dbtables']])) { 
echo "<input type=\"submit\" name=\"select_table\" value=\"Submit\"></select></form>";
mysql_free_result($result);
$table = $dbtables[0]; 
$table = trim($table); 
} 
?>

Code (markup):

this works great however further down i have used another form which basically inserts a stock code, description and price into the table that they have selected so the sql i have used is this.

$sql="INSERT INTO `$table` (item_code, item_desc, item_price)VALUES('$item_code', '$item_desc', '$item_price')";

Code (markup):

which works fine when i don't use the $table variable and type in the table automatically. so the problem is once i use the next form it resets the $table variable to nothing because they are in different forms.

Does anyone know a way around this !

so would javascript do the job for me, ot would there be a way i can insert it into the same form as my insert one

 

I think you have to do it in one form.