Question

即使在表格中创建了帐户，我也无法使用它登录，在输入新建帐户后，我收到“错误的用户名或密码”错误消息。试图运行查询，它只会在</body> </html>结束，我在这里做错了什么？也许它与数据库中的某些东西有关，请指教。

login.php文件

if($count !== 1) echo '<div class="alert alert-danger">Wrong Username or Password</div>';

index.js文件

session_start();

include("connection.php");

$missingEmail = '<p><stong>Please enter your email address!</strong></p>';
$missingPassword = '<p><stong>Please enter your password!</strong></p>'; 

if(empty($_POST["loginemail"])){
  $errors .= $missingEmail;   
    }else{
   $email = filter_var($_POST["loginemail"], FILTER_SANITIZE_EMAIL);
}
if(empty($_POST["loginpassword"])){
  $errors .= $missingPassword;   
}else{
$password = filter_var($_POST["loginpassword"], FILTER_SANITIZE_STRING);
}

if($errors){

$resultMessage = '<div class="alert alert-danger">' . $errors .'</div>';
echo $resultMessage;   
}else{

$email = mysqli_real_escape_string($link, $email);
$password = mysqli_real_escape_string($link, $password);
$password = hash('sha256', $password);
    $sql = "SELECT * FROM users WHERE email='$email' AND password='$password' AND activation='activated'";
$result = mysqli_query($link, $sql);
if(!$result){
echo '<div class="alert alert-danger">Error running the query!</div>';
exit;
}
   $count = mysqli_num_rows($result);
//If email & password don't match print error
if($count !== 1){
echo '<div class="alert alert-danger">Wrong Username or Password</div>';
}
else {
$row = mysqli_fetch_array($result, MYSQLI_ASSOC);
$_SESSION['user_id']=$row['user_id'];
$_SESSION['username']=$row['username'];
$_SESSION['email']=$row['email'];

}
}
}

Answer 1

从我可以看到你有$errors .= $missingPassword;因为$errors变量尚未创建，它会导致错误，因为你已经尝试在错误检查器上连接它。

您可以通过首先使用空值指定错误变量来修复该部分。

$errors = '';
if(empty($_POST["loginemail"])){
  $errors .= $missingEmail;   
} else {
  $email = filter_var($_POST["loginemail"], FILTER_SANITIZE_EMAIL);
}
if(empty($_POST["loginpassword"])) {
  $errors .= $missingPassword;   
} else {
  $password = filter_var($_POST["loginpassword"], FILTER_SANITIZE_STRING);
}

无法运行我的登录表单的查询

1 个答案: