Sounds like a good job for client side JavaScript. Regards,
-Josh ____________________________________ Joshua Kehn | josh.k...@gmail.com http://joshuakehn.com On Jun 7, 2011, at 12:40 AM, Chris Stinemetz wrote: > I have three drop down menus in my form. How do I make it so the > second and third menus are only visible once the prior menu was > selected? > > Below is the first two drop down menus. > > Thanks in advance. > > // Generating first menu using array. > $markets = array('MCI' => 'Kansas City', > 'STL' => 'ST. Louis', > 'ICT' => 'Wichita', > 'OMA' => 'Omaha', > 'LIN' => 'Lincoln'); > echo "<select name='term'><option value=''>Choose Market</option>\n"; > foreach ($markets as $key => $market) { > echo "<option value='$key'>$market</option>\n"; > } > echo "</select>"; > > // This will evaluate to TRUE so the text will be printed. > if (isset($markets)) { > echo "This var is set so I will print."; > } > > > > // Then, later, validating the menu > if (! array_key_exists($_POST['Market'], $choices)) { > echo "You must select a market."; > } > > $query="SELECT cell_sect FROM sector_list order by cell_sect"; > > $result = mysql_query ($query); > echo "<select name='cat'><option value=''>Choose Cell Sector</option>"; > // printing the list box select command > > while($cellSect=mysql_fetch_array($result)){//Array or records stored > in $cellSect > echo "<option value=$cellSect[cell_sect]>$cellSect[cell_sect]</option>"; > /* Option values are added by looping through the array */ > } > echo "</select>";// Closing of list box > > // This will evaluate to TRUE so the text will be printed. > if (isset($cellSect)) { > echo "This var is set so I will print."; > } > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
