Question

我有一个需要用户名和密码的登录表单。如果登录凭据错误，我希望表单顶部说“无效密码”或“无效用户名”。有人可以提供有关这样做的见解吗？

如果一个字段现在为空，则消息显示“密码无效”。我希望它有消息，即使字段中有某些内容，如果它是错误的。

这是登录表格：

 <form action="index.php?action=login" method="post">
                    <fieldset>

                        <div style="color:red;"><?php echo isset($_REQUEST['err']) && $_REQUEST['err'] == 1 ? "Invalid     Password" : "";?></div>
                        <legend>Login</legend>
                        <label for="loginName" class="required">Username:</label>
                        <input id="loginName" name="loginName" type="text"
                               value="" required />
                    <label for="password" class="required">Password:</label>
                        <input id="password" name="password" type="password"
                               value="" required />
                        <input id="submit" class="submit" type="submit" value="login"/>
                    </fieldset>
             </form>

这是登录功能（适用于会员/管理员网站，因此登录到两个帐户）：

 function connect($loginName) {
    global $db;
    $query = "SELECT email, level, password FROM members WHERE email = '$loginName'";
    $result = $db->query($query);
    $results = $result->fetch(PDO::FETCH_ASSOC);
    return $results;
 }

//Login

function login($loginName, $password) {
    $results = connect($loginName);

    if(!$results) {
        header('Location: /tire/admin/home.php?err=1');
    }

    if ($loginName === $results['email'] && password_verify($password,$results['password'])) {
        $_SESSION['loginName'] = $loginName;

        if ($results['level'] === 'a') { // 1 == Administrator
            $_SESSION['level'] = 'Administrator';
            header('Location: /tire/admin/home.php');
         } elseif ($results['level'] === 'm') { // 1 == Member
            $_SESSION['level'] = 'Member';
         header('Location: /tire/member/home.php');
         exit;
         }
     }

     header('Location: /tire/admin/home.php');
  }

//Logout
 function logout() {
    $_SESSION = array();
    session_destroy();
 }

@bakriawad这是我正在尝试你的建议但它仍然无法正常工作。它告诉我$ loginName和$ password是未定义的索引。

function error_message(){ unset($error); 
    $loginName = $_SESSION['loginName'];
{$results = connect($loginName);
    $loginName === $results['email'];

 $password = password_hash($_POST['password'], PASSWORD_BCRYPT);

  $passwords = password_verify($password,$results['password']);

   if(!$results) {$error = "Username not found";} //if no records returned,      set error to no username
   else //if found     {

     if ((isset($password)) !== (isset($passwords)))  //check password, if matched log him in
     { $error = "Password is wrong"; } //if not matched then set error message
   }
 }

   if(isset($error))echo $error; //if there is an error print it, this can      be anywhere in the page
 }

Answer 1

PHP方面：（peusedo代码）

{
  unset($error);  // or $error="";, just reset it
  $loging = select from database where username = 'username'; //get data from database
  if(!$loging) {$error = "Username not found";} //if no records returned, set error to no username
  else //if found
  {

    if ($password == $loging['pass']) {login();} //check password, if matched log him in
    else $error = "Password is wrong";  //if not matched then set error message
  }

  if(isset($error))echo $error; //if there is an error print it, this can be anywhere in the page
}

Java脚本方：

对php函数进行ajax调用，检查登录发送用户名和密码，如果正确则将页面重定向到欢迎屏幕，如果不改变框的样式和/或显示错误消息你将不得不研究这个，因为我从未使用过ajax

尝试自己动手，如果你绊倒我会乐意为你提供样品

PHP中的登录表单验证和错误消息

1 个答案: