Question

我有一个登录表单，用户可以通过该表单登录该网站。我正在检查用户名和密码是否匹配。如果匹配，则用户将被重定向到网站，但如果不匹配，则表单的上半部分将显示错误消息。

为此，我将错误消息存储在变量中，然后将其打印到表单的上半部分。但它没有获得变量的价值。代码如下：

<?php
$link = mysqli_connect("localhost", "root", "", "youtube_project");
 ?>
<!DOCTYPE html>
<html >
  <head>
    <meta charset="UTF-8">
    <title>Login Form</title>
        <link rel="stylesheet" href="css/style.css">
  </head>
  <body>
    <div class="login">
  <div class="login-triangle"></div>

  <h2 class="login-header">Log in</h2>
  <form class="login-container" name="form1" action="" method="post">
    <p style="color:red;"><?php if(isset($msg)){ echo $msg; } //Printing error msg, but couldn't get it ?></p>
    <p><input type="text" placeholder="Username" name="username" required></p>
    <p><input type="password" placeholder="Password" name="pwd" required></p>
    <p><input type="submit" name="submit1" value="Log in"></p>
  </form>
</div>
    <script src='http://cdnjs.cloudflare.com/ajax/libs/jquery/2.1.3/jquery.min.js'></script>
<?php
if(isset($_POST['submit1'])){
  $username = mysqli_real_escape_string($link, $_POST['username']);
  $pwd = mysqli_real_escape_string($link, $_POST['pwd']);
  $sql = mysqli_query($link, "SELECT * FROM admin_login WHERE username='$username' && password='$pwd'");
  if(mysqli_num_rows($sql) >= 1){
    ?>
    <script type="text/javascript">
      window.location="demo.php";
    </script>
<?php
  } else {
    $msg = "Invalid Username/Password combination"; //Storing error msg
  }
}
 ?>
  </body>
</html>

Answer 1

您需要将PHP代码移到顶部，以便设置或取消设置变量$msg。

以下是格式化的更新代码：

<?php
$link = mysqli_connect("localhost", "root", "", "youtube_project");
if(isset($_POST['submit1'])){
    $username = mysqli_real_escape_string($link, $_POST['username']);
    $pwd = mysqli_real_escape_string($link, $_POST['pwd']);
    $sql = mysqli_query($link, "SELECT * FROM admin_login WHERE username='$username' AND password='$pwd'");
    if(mysqli_num_rows($sql) >= 1){
?>
<script type="text/javascript">
    window.location="demo.php";
</script>
<?php
    } else {
        $msg = "Invalid Username/Password combination"; //Storing error msg
    }
}
?>
<!DOCTYPE html>
<html >
    <head>
        <meta charset="UTF-8">
        <title>
            Login Form
        </title>
        <link rel="stylesheet" href="css/style.css">
    </head>
    <body>
        <div class="login">
            <div class="login-triangle">
            </div>
            <h2 class="login-header">
                Log in
            </h2>
            <form class="login-container" name="form1" action="" method="post">
                <p style="color:red;">
                    <?php if(isset($msg)){ echo $msg; } //Printing error msg, but couldn't get it ?></p>
                <p>
                    <input type="text" placeholder="Username" name="username" required>
                </p>
                <p>
                    <input type="password" placeholder="Password" name="pwd" required>
                </p>
                <p>
                    <input type="submit" name="submit1" value="Log in">
                </p>
            </form>
        </div>
        <script src='http://cdnjs.cloudflare.com/ajax/libs/jquery/2.1.3/jquery.min.js'></script>
    </body>
</html>

我建议你永远不要存储纯文本密码！请使用PHP的built-in functions来处理密码安全性。

您 SQL Injections也非常开放，应该真正使用Prepared Statements。

Answer 2

你必须在显示它之前先初始化php变量。所以你必须在顶部编写php部分。

<?php
    $link = mysqli_connect("localhost", "root", "", "youtube_project");
     ?>

    <?php
    if(isset($_POST['submit1'])){
      $username = mysqli_real_escape_string($link, $_POST['username']);
      $pwd = mysqli_real_escape_string($link, $_POST['pwd']);
      $sql = mysqli_query($link, "SELECT * FROM admin_login WHERE username='$username' && password='$pwd'");
      if(mysqli_num_rows($sql) >= 1){
        ?>
        <script type="text/javascript">
          window.location="demo.php";
        </script>
    <?php
      } else {
        $msg = "Invalid Username/Password combination"; //Storing error msg
      }
    }
     ?>
    <!DOCTYPE html>
    <html >
      <head>
        <meta charset="UTF-8">
        <title>Login Form</title>
            <link rel="stylesheet" href="css/style.css">
      </head>
      <body>
        <div class="login">
      <div class="login-triangle"></div>

      <h2 class="login-header">Log in</h2>
      <form class="login-container" name="form1" action="" method="post">
        <p style="color:red;"><?php if(isset($msg)){ echo $msg; } //Printing error msg, but couldn't get it ?></p>
        <p><input type="text" placeholder="Username" name="username" required></p>
        <p><input type="password" placeholder="Password" name="pwd" required></p>
        <p><input type="submit" name="submit1" value="Log in"></p>
      </form>
    </div>
        <script src='http://cdnjs.cloudflare.com/ajax/libs/jquery/2.1.3/jquery.min.js'></script>

      </body>
    </html>

Answer 3

我希望它可以提供帮助

<?php
$link = mysqli_connect("localhost", "root", "", "youtube_project");

if(isset($_POST['submit1'])){
  $username = mysqli_real_escape_string($link, $_POST['username']);
  $pwd = mysqli_real_escape_string($link, $_POST['pwd']);
  $sql = mysqli_query($link, "SELECT * FROM admin_login WHERE username='$username' && password='$pwd'");
  if(mysqli_num_rows($sql) >= 1){
    ?>
    <script type="text/javascript">
      window.location="demo.php";
    </script>
<?php
  } else {
    $msg = "Invalid Username/Password combination"; //Storing error msg
  }
}
 ?>

<!DOCTYPE html>
<html >
  <head>
    <meta charset="UTF-8">
    <title>Login Form</title>
        <link rel="stylesheet" href="css/style.css">
  </head>
  <body>
    <div class="login">
  <div class="login-triangle"></div>

  <h2 class="login-header">Log in</h2>
  <form class="login-container" name="form1" action="" method="post">
    <p style="color:red;"><?php if(isset($msg)){ echo $msg; } //Printing error msg, but couldn't get it ?></p>
    <p><input type="text" placeholder="Username" name="username" required></p>
    <p><input type="password" placeholder="Password" name="pwd" required></p>
    <p><input type="submit" name="submit1" value="Log in"></p>
  </form>
</div>
    <script src='http://cdnjs.cloudflare.com/ajax/libs/jquery/2.1.3/jquery.min.js'></script>
  </body>
</html>

为什么我没有在以下代码中获取变量的值？

3 个答案: