Question

我一直在与代码挣扎!!我一直收到无效的密码错误，但无论多少次我调整散列的位置或在哪里检查密码我仍然得到它，即使我为不同的用户放置正确的密码...我真的需要一些帮助！我的数据库中的'用户'表是 ID 用户名密码（sha1） fullnames 手机号码用户类型这是我目前正在努力工作的代码！

       <body>
      <div data-role="page" id="login">
        <div data-role="header" data-theme="c">
          <h2 align="center"><strong>Sign In</strong></h2>
        </div>
       <div data-role="content">
          <form name="login" class="ui-corner-all" method="post" action="">
            <label for="username" class="ui-hidden-accessible">Username:</label>
            <input type="text" id="username" name ="username" placeholder="Username" required>
            <br>
            <br>
            <label for="password" class="ui-hidden-accessible">Password:</label>
            <input type="password" id="password" name ="password" placeholder="Password"  required>
            <br>
            <fieldset class="ui-grid-a">
            <div class="ui-block-a">
          <input name="login" type="submit"  id="login" formmethod="POST"          value ="login" data-ajax="false" data-theme="b">
          </div>
          <div class="ui-block-b">
          <a href="Registration.html" data-ajax="false" data-role="button"     data-theme="b">Not a user? Sign Up</a></div>
      </fieldset> 
          </form>
          <?php   
      if (isset($_POST["login"]) && !empty($_POST["login"])) {

      $user = stripslashes($_POST['username']);
      $pass = stripslashes($_POST['password']);
      $user = mysqli_real_escape_string($link, $_POST['username']);
      $pass = mysqli_real_escape_string($link, hash("SHA1",($_POST['password'])));

      if ($user && $pass) {  // CHECK ALL FIELD HAS BEEN FILLED UP

      // QUERY FROM DATABASE
      $sql = "SELECT * FROM users WHERE username='".$user."'";
      $query =mysqli_query($link,$sql) or die(mysqli_error($link));

      $numrows = mysqli_num_rows($query);

      if($numrows !== 0)
      {
      while ($row = mysqli_fetch_assoc($query))
      {
          $username = $row['username'];
          $password = $row['password'];

          }

          if($user==$username && $pass==$password)
          {
              echo "you are logged in";
              @$_SESSION['username'] = $username;
              }

              else {
                  echo "your password is incorrect";    
                  }
      }
                  else
                      die ("that user doesn't exist!");
                  }
                  else
                      die ("Please enter a username and password");
      }


          ?>

Answer 1

您的代码运行正常。我收到“你登录了”。我刚刚在mysqli_real_escape_string（）

中添加了session_start（）并替换了变量

<form method="POST" action="includes/login.php">
    <input type="email" name="email" placeholder="email" />
    <input type="password" name="password" placeholder="parola" />
    <input type="submit" value="Login" name="submit">
</form>

<?php
require_once 'config.php';

if(isset($_POST['submit'])) {
    if(!empty($_POST[email]))
    {
       $email = mysqli_real_escape_string($link,$_POST['email']);
    } 
else 
    {
        echo "Nu ati completat adresa de e-mail. <br />";
    }

if(!empty($_POST['password'])) 
    {
       $password = mysqli_real_escape_string($link,$_POST['password']);
    } 
else 
    {
        echo "Nu ati completat parola. <br />";
    }

if(!empty($_POST['email']) && !empty($_POST['password']))
{ 
    $query = ("SELECT * FROM `users` WHERE password = '".$password."' AND email = '".$email."'");
    $result = mysqli_query($link, $query);
    $row = mysqli_fetch_array($result);
    $count_rows = mysqli_num_rows($result);

    if ($count_rows == 1)
    {
            $_SESSION['login'] = "OK";
            header("Location: ../index.php");
    }

    else
    {
        header("Location: ../login.php");

    }
}}
?>

jquery移动登录表单与PHP

1 个答案: