Javascript 表格won'；t使用锚节进行验证

我在一个独立的联系人表单上有一个表单，但是当我试图在滚动页面的底部或带有锚的部分使用它时，它不会进行视觉验证。换句话说，错误消息不会出现。如果填写了所有必填字段，表单将发送一封电子邮件。但从视觉上看什么都没有发生！？

要了解它应该如何工作，请查看一个独立的contact.php patrickmchugh.com/test/contact.php这是错误消息应该如何工作的

<?php   
// check for form submission - if it doesn't exist then send back to contact form  
if (!isset($_POST['save']) || $_POST['save'] != 'contact') { 
    header('Location: index.php#content'); exit; 
} 

// get the posted data 
$name = $_POST['contact_name']; 
$email_address = $_POST['contact_email']; 
$subject = $_POST['contact_subject'];
$message = $_POST['contact_message']; 

// check that a name was entered 
if (empty($name)) 
    $error = 'You must enter your name.'; 
// check that an email address was entered 
elseif (empty($email_address))  
    $error = 'You must enter your email address.'; 
// check for a valid email address 
elseif (!preg_match('/^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,3})$/', $email_address)) 
    $error = 'You must enter a valid email address.'; 
// check that a message was entered 
elseif (empty($message)) 
    $error = 'You must enter a message.'; 

// check if an error was found - if there was, send the user back to the form 
if (isset($error)) { 
    header('Location: index.php?e='.urlencode($error).'#content2'); exit;
} 

// write the email content 
$email_content = "Name: $name\n"; 
$email_content .= "Email Address: $email_address\n"; 
$email_content .= "Subject: $subject\n";
$email_content .= "Message:\n\n$message"; 

// send the email 
mail ("patrick@patrickmchugh.com", "Enquiry from Connolly O'Neill Website", $email_content); 

// send the user back to the form 
header('Location: index.php?s='.urlencode('Thank you for your message.').'#contact2'); exit; 

---

提交表单后，您的url将是
？e=您+必须+输入+您的+姓名。#content2
因此
您需要在表单页面上获取并回显错误，如：-

if(isset($_GET['e'])) {
  echo $_GET['e'];  // or echo urldecode($_GET['e']);
}

成功提交后的相同，如：-

if(isset($_GET['s'])) {
  echo $_GET['s'];  // or echo urldecode($_GET['s']);
}

Rakesh Sharma我不明白，我是在交换代码还是只是添加这个？把我的代码放在你想显示错误和消息的地方，然后看到这样的结果？？//如果（空（$name））$error=“您必须输入您的姓名”，请检查是否输入了姓名；如果（isset（$_-GET['e']）{echo$_-GET['e']；//或echo-urldecode（$_-GET['e']）}仅使用给定的代码，如抱歉，我将其放入php代码中。我意识到现在我需要把它放到索引文件中。非常感谢。谢谢