Home  >  Article  >  Backend Development  >  php判断正常访问和外部访问的示例_PHP

php判断正常访问和外部访问的示例_PHP

WBOY
WBOYOriginal
2016-06-01 11:57:19851browse

php判断正常访问和外部访问
复制代码 代码如下:
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;
?>





Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn