【发布时间】:2017-08-24 15:55:53
【问题描述】:
我创建了一个 html 表单,旨在做两件事 - 将表单信息提交到指定的电子邮件地址并将用户重定向到网站上的另一个页面。我创建了一个 .php 表单来管理这些操作。表单成功重定向并发送了一封电子邮件。但是,电子邮件具有正确的目的地、主题和消息,但未能提取用户提交给表单的信息 - 即姓名、地址、电子邮件等。
我对 Web 开发完全陌生,但如果您能提供任何见解,我将不胜感激。谢谢!
HTML:
<section>
<div class="row">
<div class="col-lg-offset-3 col-lg-6 col-md-12"><form name="Whitepaper
Request form" method="POST" action="Info-Request.php" class="text-center">
<input name="First_Name" type="text" required class="contact-field-sm"
id="fname" placeholder="First Name">
<input name="Last_Name" type="text" required class="contact-field-sm"
id="lname" placeholder="Last Name"><br><br>
<input name="Address" type="text" class="contact-field-sm" id="address"
placeholder="Address">
<input name="City" type="text" class="contact-field-sm" id="city"
placeholder="City"><br><br>
<input name="State" type="text" class="contact-field-sm" id="state"
placeholder="State">
<input name="Zip" type="text" class="contact-field-sm" id="zip"
placeholder="Zip"><br><br>
<input name="Phone" type="tel" class="contact-field-sm" id="phone"
placeholder="Phone">
<input name="Email" type="email" class="contact-field-sm" id="email"
placeholder="Email"><br><br>
<input name="Company" type="text" required class="contact-field-sm"
id="company" placeholder="Company">
<input name="Title" type="text" required class="contact-field-sm" id="title"
placeholder="Title"><br><br>
<input id="submit" type="submit" value="Submit">
</form>
PHP:
<?php
if(!isset($_POST['submit']))
$to = "email@example.com";
$from = 'Whitepaper Request Form';
$email_subject = " ";
$email_message = "Results from Whitepaper Request Form:\n\n";
function clean_string($string) {
$bad = array("content-type","bcc:","to:","cc:","href");
return str_replace($bad,"",$string);
}
$First_Name = $_POST['First_Name'];
$Last_Name = $_POST['Last_Name'];
$email_message .= "First Name: ".clean_string($First_Name)."\n";
$email_message .= "Last Name: ".clean_string($Last_Name)."\n";
$email_message .= "Address: ".clean_string($Address)."\n";
$email_message .= "City: ".clean_string($City)."\n";
$email_message .= "State: ".clean_string($State)."\n";
$email_message .= "Zip: ".clean_string($Zip)."\n";
$email_message .= "Phone: ".clean_string($Phone)."\n";
$email_message .= "Email: ".clean_string($Email)."\n";
$email_message .= "Company: ".clean_string($Company)."\n";
$email_message .= "Title: ".clean_string($Title)."\n";
mail($to,$from,$email_subject,$email_message);
header('Location: Example_Retro_WhtP.pdf');
?>
[编辑] PHP:
<?php
if(isset($_POST['submit'])){
$to = 'example@example.com';
$subject = 'Whitepaper Request Form';
$email_message = 'Results from Whitepaper Request Form:\n\n';
$headers = 'From: example@example.com';
function clean_string($string) {
$bad = array("content-type","bcc:","to:","cc:","href");
return str_replace($bad,"",$string);
}
$email_message .= "First Name: ".clean_string($_POST['First_Name'])."\n";
$email_message .= "Last Name: ".clean_string($Last_Name)."\n";
$email_message .= "Address: ".clean_string($Address)."\n";
$email_message .= "City: ".clean_string($City)."\n";
$email_message .= "State: ".clean_string($State)."\n";
$email_message .= "Zip: ".clean_string($Zip)."\n";
$email_message .= "Phone: ".clean_string($Phone)."\n";
$email_message .= "Email: ".clean_string($Email)."\n";
$email_message .= "Company: ".clean_string($Company)."\n";
$email_message .= "Title: ".clean_string($Title)."\n";
mail($to,$subject,$email_message,$headers);
header('Location: Example_Retro_WhtP.pdf');
exit;
}
?>
【问题讨论】:
-
名字和姓氏应该通过,但没有其他信息,因为您从不引用 POST 变量。例如。
$_POST['Email'] -
名字和姓氏作为 POST 变量也不起作用。
-
您已经对名字和姓氏做了正确的事情(仅)所有其他表单变量都没有包含在
$_POST['']中。尝试在您的 PHP 脚本中添加print_r($_POST)或var_dump()以查看表单中的内容。 -
@j08691,我不知道你说的
print_r($_POST)是什么意思 -
在
<?php之后将该行添加到您的PHP脚本中
标签: php html forms html-email