这是员工的注册表单。如果员工是临时员工或永久员工，则必须签入单选按钮。如果是临时的，他应该填写输入的合同编号，如果是永久的，则应填写雇用日期。如何将单选按钮与输入绑定，以便他不能“交叉完成”，例如，如果他是临时的，请填写招聘日期。我想让他只有在按下特定的单选按钮时才能填写每个输入。
代码：
<?php include 'dbconfig.php';?>
<?php header('Content-Type: text/html; charset=utf-8');?>
<!DOCTYPE HTML PUCLIC "-//W3C//DTDHTML
4.0 Transitional//EN"><HTML>
<HEAD>
<link rel="stylesheet" type="text/css" href="logintest.css">
<meta http-equiv="content-type" content="text/html; charset=UTF-8">
</HEAD>
<button class="btn" TYPE="submit" name="goback" onclick="window.location.href='login.php'">Go Back </button>
<?php
error_reporting(E_ALL); ini_set('display_errors', 1);
if(isset($_POST['submit']))
{
$sql = "INSERT INTO employee (empID,EFirst,ELast,username, passcode)
VALUES ('".$_POST["empID"]."','".$_POST["EFirst"]."','".$_POST["ELast"]."','".$_POST["username"]."','".$_POST["passcode"]."')";
$result = mysqli_query($conn,$sql);
$answer=$_POST['kind'];
if($answer='permament'){
$sql1 = "INSERT INTO permament_employee (empID,Hiring_Date) VALUES ('".$_POST["empID"]."','".$_POST["date"]."')";
$result1 = mysqli_query($conn,$sql1);
}
if ($answer='temporary'){
$sql2= "INSERT INTO temporary_employee(empID) VALUES ('".$_POST["empID"]."')";
$result2 = mysqli_query($conn,$sql2);
}
echo "<script> location.replace('login.php') </script>";
}
不负相思意