这就是我所做的,

HTML

<select name="gender">
<option value="select">  Select </option>
<option value="male">    Male   </option>
<option value="female">  Female </option>
</select>

PHP脚本

$Gender  = $_POST["gender"];

但我得到这些错误

Notice: Undefined index: gender in C:\xampp\htdocs\omnama\signup.PHP on line 7

PHP脚本

$Gender  = isset($_POST["gender"]); ' IT returns a empty string ? why ?

HTML

<form name="signup_form"  action="./signup.PHP" onsubmit="return validateForm()"   method="post">
<table> 
  <tr> <td> First Name    </td><td> <input tyPE="text" name="fname" size=10/></td></tr>
  <tr> <td> Last Name     </td><td> <input type="text" name="lname" size=10/></td></tr>
  <tr> <td> Your Email    </td><td> <input type="text" name="email" size=10/></td></tr>
  <tr> <td> Re-type Email </td><td> <input type="text" name="remail"size=10/></td></tr>
  <tr> <td> Password      </td><td> <input type="password" name="paswod" size=10/> </td></tr>
  <tr> <td> Gender        </td><td> <select name="gender">
  <option>                Select </option>    
  <option value="male">   Male   </option>
  <option value="female"> Female </option></select></td></tr> 
  <tr> <td> <input type="submit" value="Sign up" id="signup"/> </td> </tr>
 </table>
 </form>

这是我的PHP脚本

<?PHP
  $con     = MysqL_connect("localhost","root","");
  $fname   = $_POST["fname"];
  $lname   = $_POST["lname"];
  $email   =  $_POST["email"];
  $paswod  = $_POST["paswod"];
  $Gender  = $_POST["gender"];
  MysqL_select_db("homepage");

  if(MysqL_num_rows(MysqL_query("SELECT Email From users WHERE Email = '$email'",$con)))
  {
  echo "userid is already there";
  }
  else
  { 
  $sql= "INSERT INTO users (FirstName,LastName,Email,Password,Gender)
  VALUES
  ('$fname','$lname','$email','$paswod','$Gender')";

  if (!MysqL_query($sql,$con))
  {
  die('Error: ' . MysqL_error());
  }
  echo "created";
  }
 ?>

请帮帮我.我必须在PHP中获取所选的索引值.

我已阅读此link以使用< select>用PHP标记.