Making a Select Menu from MySQL ID and display results ...

[TylerNZ]

Ok, so I've been LOOKING and LOOKING ... and I've found bits and pieces ... but here's what I've got and what I want to accomplish.

I have a MySQL database, with one table "staff", and in that I have about 10 fields with information about the staff member. And about 20 rows of dif. staff members.

I have managed to make an Index PHP page that pulls just the name and email. And make the name link thru to a detailed page which has the rest of the information. I've even got some Previous and Next links working!

HOWEVER, now I want to go the next step and create a page that Staff Members can access to change their own profiles.

So 1st) I need to have an Admin Passworded Page (that shouldn't be hard, I've seen lots of those around)

2nd) I need to have a select menu (dropdown) that when users click on a Staff Name their details are shown and editable.

3rd) Sending the updated data back to the database.

I know the 3rd part is easy to find as well, but I'm REALLY stuck on the 2nd thing. How can I make a dropdown (select) menu pulling from my DB? And then have it change the details on the page?

Thanks!!
Joel

Oh, here's the code I have so far:

<form name=wordlist method=post action="<?php echo $PHP_SELF?>"><center> 
Select a Word Type or Enter a New Type<BR> 
<input type="hidden" value="id"> 
<select name="selecttype" class="sel"><option>Staff Member</option> 
<? 
{     

   $result = mysql_query("SELECT * FROM staff ORDER BY StaffID ASC"); 
    if ($myrow = mysql_fetch_array($result)) { 
      // display list if there are records to display 
      do { 
        printf("<option value=\"%s\">%s</option>", $myrow["selecttype"], $myrow["selecttype"]); 

  } while ($myrow = mysql_fetch_array($result)); 

  } 
} ?></select>

[bastien]

shouldn't the staff member only have access to his/her own page? for security purposes? otherwise someone might play tricks on staff members?

<?  

{      

   $result = mysql_query("SELECT * FROM staff ORDER BY StaffID ASC");  

    if (mysql_num_rows($result)>0) {  

      // display list if there are records to display 
       while ($myrow = mysql_fetch_array($result)){  

        printf("<option value=\"".$myrow["selecttype"]."\">".$myrow["selecttype"] );  

  }  
} 

?>