php判断正常访问和外部访问的示例

  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">