Drop down boxes from db

[corruption]

I'm writing a script to update the positions of people on an executive board.

I have a page with all of the positions set up as links. The links go to a page where i want the position to show up and then the list of brothers show up in a drop down menu.

this is the script that i have currently:

<table class="content_table">
<tr><td colspan="2">Executive Board Members</td></tr>
 <tr><td> <? //select the office chosen on the previous page
   $office_id = $_GET['office_id'];
   $sql = "SELECT * from exec where office_id = $office_id";
   $query = mysql_query($sql) or die(mysql_error());
   if(mysql_num_rows($query) > 0)
    {
    ?>
    <?php 
        while($row = mysql_fetch_array($query)) 
    {
	$position = $row['position'];
	//create the form to edit the officers
	echo "<form method=\"post\" action=\"updateexec.php\">";
	echo "<td>Position:</td><td><input type=\"text\" name =\"position\" value='$position' /></td>";
	 }
	 } 
    else
    {
    echo "No rows returned!";
    }
	 //get the list of brothers
	 echo "<td>Brother:</td>";
	 echo "<td><select name='brother'></td>";
	 $sql2 = "SELECT fname,lname,brother_id from brothers";
	 $brothers = mysql_query($sql2) or die(mysql_error());
	 if (mysql_num_rows($brothers) > 0)
	 {
	 while($row2 = mysql_fetch_array($brothers))
	 {
	 //put the list of brothers in a drop down menu for me to select and edit as needed.

 echo "<td><option value=".$row2['brother_id'].">".$row2['fname']." ".$row2['lname']."</option>";
 echo "</select></td>";

} 

}

?>

[drawmack]

You can't put each option in a table cell, that doesn't even make any sense. If your problem is something other then this please elaborate on what the problem actually is.

[corruption]

I don't want each option in a new table cell. For that problem then i'd guess i have a problem with where things are in the loops. I need the options to actually show up in the drop down menu. If that problem also is in the loop structure then i'll have to take a look at it.

[Roger_Ramjet]

You may not want each item in a table data cell, but that is what you have coded, as drawmack pointed out.

while($row2 = mysql_fetch_array($brothers)) 
     { 
     //put the list of brothers in a drop down menu for me to select and edit as needed. 

 echo "<td><option value=".$row2['brother_id'].">".$row2['fname']." ".$row2['lname']."</option>"; 
 echo "</select></td>"; 

}

Move the </select></td> outside of the while clause:

} 
     //get the list of brothers 
	$sql2 = "SELECT fname,lname,brother_id from brothers"; 
	$brothers = mysql_query($sql2) or die(mysql_error()); 
    if (mysql_num_rows($brothers) > 0)  {
    	//put the list of brothers in a drop down menu for me to select and edit as needed. 
    	echo "<td>Brother:</td>";
    	echo "<td><select name='brother'></td>";
    	while($row2 = mysql_fetch_array($brothers)) {
    		echo "<option value=".$row2['brother_id'].">".$row2['fname']." ".$row2['lname']."</option>";
    	}
    	echo "</select></td>"; 
	}

or something like that.