help: another dropdown menu problem

[fibonacci]

hi all,

i need a help to fix this problem tho..
here is my code

<label for="title">Title:</label>
            <select name="title">
    <?php
            echo "title:".$listAdmin[0]['title'];
         $title = array('Mr.','Ms.','Mrs.');
         foreach ($title as $keytitle){
            if($listAdmin[0]['title']==$keytitle)

            $selected="selected";
        else
            $selected=" ";
  }
  printf("<option value='%s'%s>%s</option>",$keytitle,$selected,$keytitle);
?>
        </select>

i'm doing an edit page for admin. but when to display a dropdown menu it doesn't work as it should be. it just display the exact title from the database instead of all title listed.

can anybody help me?
thanks..

[NogDog]

Looks like you need to move the closing "}" to after the printf() command.

[leatherback]

Also, -although not required here- just for good coding practice: Get used to using {} consistently:

if($listAdmin[0]['title']==$keytitle)
  {       

   $selected="selected";
  }
else
  {
  $selected=" "; 
  }