php判断正常访问和外部访问的示例
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2024-01-04 18:53:23
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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;
?>