菜鸟问题：传值问题

Question

rizhisec.action = "rizhisec.php?action=ready&slt=" + document.getElementById("slt").value + "&inpt=" + document.getElementById("inpt").value;

Answer 1

还有我下拉框选择的是通过数据库查询用户表中的姓名，我想把姓名对应的用户ｉｄ也通过ｕｒｌ传递过去。怎样才能在（提交前）根据选择的用户名在用户表中查出对应的用户ｉｄ。即提交前我怎么才能取下拉框中的值。
以下是我用用户中姓名填充下拉列表的代码：<select size="1" name="sec1" style="font-size: 9pt; font-family: serif">
                                                                <?
                                                                $connect = mysql_pconnect("localhost","root","hcedagb");
                                                                mysql_select_db("icbcoa",$connect);
                                                                $sql = "select * from oa_user";
                                                                 mysql_query("SET NAMES 'GBK'");
                                                                $result = mysql_query($sql);
                                                                while($data=mysql_fetch_assoc($result))
                                                                {
                                                                $name=$data['name'];
                                                                ?>
                                                                <option><?=$name?></option>
                                                                <?
                                                                  }
                                                                 mysql_close($connect);
                                                                ?>
                                                                </select></td>

Answer 2

关注一下，很久没碰PHP了，这个问题还是比较常见

Answer 3

<option value=<?php echo $data['id'];?>><?=$name?></option>