Hello All, I have searched far and wide and I cannot find the exact answer to the question I have. The issue is this, I need to ultimately pass an 'id' and from an 'item' in my db (mysql) to the browser when a user makes a selection from a menu I have. Currently I am able to populate the menu with my db items just fine, however, the best I can get this script to do is recognize the first selection! Any selection after that doesn't work at all. As I am testing right now, I simply print out the id and name of the selection from the menu on the same screen, right below the drop down menu. I am currently using the following code with no luck: $sql="SELECT id, name FROM items WHERE something = 'No' ORDER BY name asc"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $id=$row["id"]; $name=$row["name"]; $options.="<OPTION VALUE=\"$id/testpage.html\">".$name; } echo " // There is more code before this form but I have left it out for readability // <form action=\"GoToLocation.cgi\"> <select id=\"newLocation\"> <option selected=\"selected\">Select an Item</option> $options </select> <noscript> <input type=\"submit\" value=\"Go There!\" /> </noscript> </form> "; PHP: The following is the javascript code from file (script01.js): window.onload = initForm; window.onunload = function() {}; function initForm() { document.getElementById("newLocation").selectedIndex = 0; document.getElementById("newLocation").onchange = jumpPage; } function jumpPage() { var newLoc = document.getElementById("newLocation"); var newPage = newLoc.options[newLoc.selectedIndex].value; if (newPage != "") { window.location = newPage; } } Code (markup): I am definitely doing something wrong but I can't figure it out for the life of me.
Got it working!! Not sure if anyone is interested but I had to change: from this $options.="<OPTION VALUE=\"$id/testpage.html\">".$name; Code (markup): to this $options.="<OPTION VALUE=\"http://root_of_website/$id/testpage.html\">".$name; Code (markup):