如何在下拉菜单中选择查询结果

Question

I have this database:

  **SBU                **SBU_Users
 *id | *SBU_Name       *users_id | *sbu_id
  1  |  SBU1               1     |   3
     : 
  4  |  SBU4

And this my code:

$query = "SELECT sbu.SBU_Name, sbu_users.sbu_id FROM sbu, sbu_users WHERE sbu_users.sbu_id=sbu.id AND sbu_users.users_id=".$users_id."";
        $result = mysql_query($query, $db) or die (mysql_error($db));
            while ($row = mysql_fetch_array($result))
            {
            extract($row);

            }
            echo "<select name=sbuuser>";
            echo "<option value='1' ";if ($sbu_id==1){echo "selected";} else {echo "";}echo ">SBU1</option>";
            echo "<option value='2' ";if ($sbu_id==2){echo "selected";} else {echo "";}echo ">SBU2</option>";
            echo "<option value='3' ";if ($sbu_id==3){echo "selected";} else {echo "";}echo ">SBU3</option>";
            echo "<option value='4' ";if ($sbu_id==4){echo "selected";} else {echo "";}echo ">SBU4</option>";

I have a problem in selecting the right SBU_Name, because my code keeps selecting the last SBU_Name. Please help me.

Answer 1

Maybe something like this?

$query = "SELECT sbu.SBU_Name, sbu_users.sbu_id FROM sbu, sbu_users WHERE sbu_users.sbu_id=sbu.id AND sbu_users.users_id=".$users_id."";
        $result = mysql_query($query, $db) or die (mysql_error($db));

            echo "<select name=sbuuser>";

            while ($row = mysql_fetch_array($result))
            {
            extract($row);

            echo "<option value='1' ";if ($sbu_id==1){echo "selected";} else {echo "";}echo ">SBU1</option>";
            echo "<option value='2' ";if ($sbu_id==2){echo "selected";} else {echo "";}echo ">SBU2</option>";
            echo "<option value='3' ";if ($sbu_id==3){echo "selected";} else {echo "";}echo ">SBU3</option>";
            echo "<option value='4' ";if ($sbu_id==4){echo "selected";} else {echo "";}echo ">SBU4</option>";
            }

            echo "</select>";