php-如何防止自定义发布请求到表单的“操作”页面
作者:互联网
我正在制作一个表单,我想使提交PHP页面仅在表单提交后才可访问,以防止对我的PHP页面进行自定义请求.
这是我的form.html:
<html>
<head>
<title>Name/Surname form</title>
</head>
<body>
<form id="form1" method="POST" action="processData.php">
Name: <input type="text" id="name" name="name"><br>
Surname: <input type="text" id="surname" name="surname"><br>
<input type="submit" value="Submit form">
</form>
</body>
</html>
然后是我的processData.php:
<?php
if(!isset($_POST['name'],$_POST['surname'])) die;
include ("config.php");
//connect
$mysqli = new mysqli($dbhost, $dbuser, $dbpassword, $dbname); //variables from config.php
//check connection
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
if ($stmt = $mysqli->prepare("INSERT INTO name_surname_table (name, surname) values (?, ?)")) {
//bind
$stmt->bind_param('ss', $name, $surname);
//set
$name=$_POST['name'];
$surname=$_POST['surname'];
//execute
$stmt->execute();
//close
$stmt->close();
}
else {
//error
printf("Prepared Statement Error: %s\n", $mysqli->error);
}
?>
问题是,如果我在不提交上一页中的表单的情况下进行自定义发布请求,则数据将提交到db,这意味着自动化程序可以将所需的任何内容都放入db中.如何防止这个?
谢谢!
解决方法:
您有很多选择可以防止机械手自动提交表单:
>使用Captcha或一些简单的计算字段(例如,给出3 2的是什么?)
>带令牌
>用$_REQUEST变量进行检查
令牌可以像这样完成:
session_start();
//generate a unique string
$token = uniqid(rand(), true);
//Store the token
$_SESSION['token'] = $token;
//Store the token_time
$_SESSION['token_time'] = time();
向表单添加隐藏的输入:
<input type="hidden" name="token" id="token" value="<?php echo $token;?>"/>
并且比将表单提交到数据库之前:
session_start();
//If token exists
if(isset($_SESSION['token']) && isset($_SESSION['token_time']) && isset($_POST['token']))
{
//If it's the same as the posted one
if($_SESSION['token'] == $_POST['token'])
{
//For example 15 mins ago
$old_time = time() - (15*60);
//If token hasn't expired
if($_SESSION['token_time'] >= $old_time)
{
//Do your queries here
}
}
}
标签:form-submit,html,php 来源: https://codeday.me/bug/20191031/1973351.html