Question

我正在通过在线教程创建我的第一个登录/注册功能到我的网站。每件事似乎都运行良好，我的问题是在教程中，在索引页面中设置了php if ($_SERVER['REQUEST_METHOD'] == 'POST')，该页面检查是否所有字段然后将它们插入到数据库中。但对我来说这似乎不起作用。但是，如果我将代码放在表单操作重定向的页面上，它可以正常工作。这是正确的方法吗？我不想这么想，因为我想在继续之前检查所有变量。

因此，如果有人愿意教育我，那就太棒了。

这是我的PHP代码仍然没有完全完成，但我想先清除它。

include方法

使用它

<?php 

    //setup some variables/arrays
    $action = array();
    $action['result'] = null;

    //check if the form has been submitted
    if ($_SERVER['REQUEST_METHOD'] == 'POST'){

       $message = "wrong answer";
       echo "<script type='text/javascript'>alert('$message');</script>"; 


      $firstName = mysqli_real_escape_string($link,$_POST['firstName']);
      $lastName = mysqli_real_escape_string($link,$_POST['lastName']);
      $password = mysqli_real_escape_string($link,$_POST['sign-up-password']);
      $confirmPassword = mysqli_real_escape_string($link,$_POST['password-confirm']);
      $email = mysqli_real_escape_string($link,$_POST['email2']);

      //quick/simple validation
      if(empty($firstName)){ $action['result'] = 'error';}
      if(empty($lastName)){ $action['result'] = 'error';}
      if(empty($password)){ $action['result'] = 'error';}
      if(empty($email)){ $action['result'] = 'error';}

      if($password != $confirmPassword){ $action['result'] = 'error';}

      if($action['result'] != 'error'){

          $add = mysqli_query($link,"INSERT INTO `users` VALUES(NULL,'$firstName','$lastName','$password','$email',0)");


          if($add){
              //the user was added to the database    

              //get the new user id
              $userid = mysqli_insert_id($link);

              //create a random key
              $key = $firstName . $email . date('mY');
              $key = md5($key);

              //add confirm row
              $confirm = mysqli_query($link,"INSERT INTO `confirm` VALUES(NULL,'$userid','$key','$email')"); 

              if($confirm){

                  //let's send the email
              }
          }else{

              $action['result'] = 'error';
              array_push($text,'User could not be added to the database. Reason: ' . mysql_error());

          }
      }else{

      }
    }

  ?>

PHP注册代码

0 个答案: