Question

我想通过刷新防止双重插入但我所做的并不起作用！我哪里错了？

<?php
  session_start();
  $form_token = uniqid();
  $_SESSION['user_token'] = $form_token;
?>
<form method="post">
  <input type="hidden" name="user_token" value="<?php echo  $_SESSION['user_token'];  ?>" />
  <input type="submit" name="submit" value="Submit">
</form>
<?php
  if (isset($_POST['submit'])){
  if ($_POST['user_token'] == $_SESSION['user_token']) {
     echo "TEST OK";
     unset($_SESSION['user_token']);
   } else {echo "Session Expired";}
}
?>

Answer 1

您需要对代码的顺序进行一些修改。

<?php 
       session_start();
    // If this is a post request, you don't need to create a new session. Just check for the eixsting value.
    if (isset($_POST['submit']))
    {
        if($_SESSION['user_token'] == $_POST['user_token']) {
            echo "TEST OK";
        } else { 
          echo "Session expired";
        }
    } else{ // Start a new session ONLY if this is the first time the user is entering this page.
       $form_token = uniqid();
       $_SESSION['user_token'] = $form_token;
    }

    include_once('./config.php');
?>
    <form method="post">
    <input type="hidden" name="user_token" value="<?php echo $_SESSION['user_token'];?>" />
    <input type="submit" name="submit" value="Submit">
    </form>

注意：此代码块仅用于帮助您找出双重插入问题（或者更确切地说是始终过期的会话＆＃39;输出）。但是，这可能不是在PHP中实现会话的最佳方式。您可能希望在PHP Sessions上阅读更多内容。查看官方文档注释中给出的示例。这应该可以让你对自己的成就有所了解，

Answer 2

感谢DeadLock，我以这种方式解决了

session_start();
if (isset($_POST['submit']))
    {
    if($_SESSION['user_token'] == $_POST['user_token']) {
        unset($_SESSION['user_token']);
        $form_token = uniqid();
        $_SESSION['user_token'] = $form_token;
        include_once('./arrivi.php');
        include_once('./script/controlloarrivi.php');
        } else {
       header("location: ./2.php");
      }
    } else{
    $form_token = uniqid();
    $_SESSION['user_token'] = $form_token;
    include_once('./arrivi.php');
    }
?>

现在我有另一个问题！我需要重新加载include_once（'。/ arrivi.php'）;当我按提交但我有结果 include_once（ './脚本/ controlloarrivi.php'）;

我怎么能这样做？我读到我需要使用ajax！但是如何？

如何使用令牌进行唯一插入

2 个答案: