php判断正常访问和外部访问的示例_php技巧

php判断正常访问和外部访问

复制代码 代码如下:

<?php
session_start();
if(isset($_POST['check'])&&!empty($_POST['name'])){
if($_POST['check'] == $_SESSION['check']){
echo "正常访问";
}else{
echo "外部访问";
}
}
$token = md5(uniqid(rand(),true));
$_SESSION['check'] = $token;
?>
<form method="post" action="">
<input type="text" name="name"/>
<input type="text" name="check" value="<?=$token;?>">
<input type="submit">