Question

我和我的伙伴已经连续工作了近24小时，试图解决这个问题，我觉得这很简单，我们不应该有问题。我们正在尝试使用sqlite数据库中的PHP对用户进行身份验证。我们已经看过无数的帖子和教程，关于如何使用mySQL数据库，但不是sqlite3。我们确实找到的 one 就在这里，它对我们不起作用（不，我们不仅仅是复制和粘贴）。那么，谁能告诉我们出了什么问题？我们不断收到服务器500错误，而且根本没有验证。首先，php将显示在页面上，但现在我们只是得到错误。简而言之 - 我们试图从sqlite数据库中提取电子邮件地址和密码信息并使用php验证它，因此用户可以登录并查看成员主页。 PHP代码：

    <?php
$dir = 'sqlite:db/Quizionality.db':
 $db = new PDO($dir) or die ("Unable to open");


 $email_address = $_POST['email_address'];
 $password = $_POST['password'];

 $email_address =($email_address);
 $password =($password);

 $email_address = intval($_POST["email_address"]);
 $password = intval($_POST["password"]);
 $sql = "SELECT * FROM user_info WHERE email_address='$email_address' AND password='$password'";
 $sql .= "where email_address=? AND where password=?";

 $stmt = $db->prepare($sql);
 $stmt->bindParam(1, $email_address);
 $stmt->bindParam(2, $password);
 $result = $stmt->execute($sql);

 while($column=$stmt->fetch())
 {
     echo $column["email_address"];
     echo $column["password"];
 };
  };


 if($email_address==1){
     echo "<h1>Logged In</h1>";
 };

?>

HTML

<!DOCTYPE html> 
<html>
  <head>
    <title>Quizonality</title>

    <link rel="stylesheet" href="css/QuizonalityLogInForMembers.css" type="text/css" />
  </head>
  <body>
    <header>
      <nav id="header_wrapper">
        <h1 id="logo"><a href="#"> QUIZONALITY</a></h1>
        <ul id="navigate">
          <li><a href="#">Home</a></li>
          <li><a href="#">Contact</a></li>
          <li><a href="#">Login</a></li>
          <li><a href="#">Create</a></li>
        </ul>
      </nav>
    </header>


<body>

<div id="main">


<h2 id="main_head"> Already Existing Members </h2>

<form id="loginform" name="myForm" method="post" action="login_for_mems.php">

<div>Username (Email):</div>
<input  id="username" type="text" name="username"  >


<div>Password:</div>
<input id="password" type="text" name="password" type="password">

<br>
<input type="submit" value="Login" >
</br>

</div>

</form>




 <footer>
   <div id="footer_nav">
    <div id="footer_nav_wrapper">

     <ul>
      <li><a href="#">Home</a></li>
      <li><a href="#">Contact</a></li>
      <li><a href="#">Login</a></li>
      <li><a href="#">Create</a></li>
     </ul>



  </footer>
 </body>
</html>

提前感谢您的任何建议。

** PHP UPDATE w /错误报告 - 这是我们目前的

    <?php
error_reporting(-1);
ini_set('display_errors', 'On');

$email = isset($_POST['username'])? $_POST['username'] : ''; 
$password = isset($_POST['password'])? $_POST['password'] : '';

$db = new PDO('sqlite:db/Quizionality.db');

$stmt = $db->prepare('SELECT * FROM user_info WHERE email_address = ? and password = ?');


$stmt->execute(array($email, $password));

if($stmt->rowCount())
    {
    echo "<h3>Logged In</h3>";
    }
  else
  {
    echo "<h3>Login Failed</h3>";
  }
?>

Answer 1

请注意您的问题建议您将密码保存为数据库中的纯文本 - 您从不想要这样做（使用哈希代替）但以下代码旨在与您现有的数据库一起使用：

//get email and pass from POST - set to empty string if not set
$email = isset($_POST['username'])? $_POST['username'] : ''; //Your form html uses name="username" so thats the key we use here
$password = isset($_POST['password'])? $_POST['password'] : '';

//create PDO object with sqlite db
$db = new PDO('sqlite:db/Quizionality.db');

//prepare statement, with unnamed query parameters denoted by ? placeholder
$stmt = $db->prepare('SELECT * FROM user_info WHERE email_address = ? and password = ?');

//bind query parameters to placeholders and execute query
$stmt->execute(array($email, $password));

//rowCount does not work for SQLITE, however empty array is 'falsy', an array with content (in the case of a successfull login) is 'truthy', so:
if($stmt->fetchAll()){
    //do something now user is authenticated, eg set a SESSION value, redirect etc
    echo 'Logged In';
}

同样，我强烈建议您更改数据库以使用密码哈希，请参阅http://php.net/manual/en/function.password-hash.php

此外，您的代码中的许多问题都是明显的sytax错误 - 任何体面的PHP IDE都应该能够突出显示这些 - 所以我建议您使用一个。 Netbeans是一个很好的免费的，PHPStorm是一个（可以说是更好的）付费的

使用Sqlite数据库中的PHP验证用户登录

1 个答案: