为什么login.php一直显示无效密码？
为什么login.php一直显示无效密码？
提问于 2019-02-22 00:14:55
因此，我有一个登录页面(login.php)
<?php
// Initialize the session
session_start();
// Check if the user is already logged in, if yes then redirect him to welcome page
if(isset($_SESSION["loggedin"]) && $_SESSION["loggedin"] === true){
header("location: asdasd.php");
exit;
}
// Include config file
require_once "config.php";
// Define variables and initialize with empty values
$username = $password = "";
$username_err = $password_err = "";
// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST"){
// Check if username is empty
if(empty(trim($_POST["username"]))){
$username_err = "Please enter username.";
} else{
$username = trim($_POST["username"]);
}
// Check if password is empty
if(empty(trim($_POST["password"]))){
$password_err = "Please enter your password.";
} else{
$password = trim($_POST["password"]);
}
// Validate credentials
if(empty($username_err) && empty($password_err)){
// Prepare a select statement
$sql = "SELECT id, felhasznalonev, jelszo FROM diakok WHERE felhasznalonev = ?";
if($stmt = mysqli_prepare($link, $sql)){
// Bind variables to the prepared statement as parameters
mysqli_stmt_bind_param($stmt, "s", $param_username);
// Set parameters
$param_username = $username;
// Attempt to execute the prepared statement
if(mysqli_stmt_execute($stmt)){
// Store result
mysqli_stmt_store_result($stmt);
// Check if username exists, if yes then verify password
if(mysqli_stmt_num_rows($stmt) == 1){
// Bind result variables
mysqli_stmt_bind_result($stmt, $id, $username, $hashed_password);
if(mysqli_stmt_fetch($stmt)){
if(password_verify($password, $hashed_password)){
// Password is correct, so start a new session
session_start();
// Store data in session variables
$_SESSION["loggedin"] = true;
$_SESSION["id"] = $id;
$_SESSION["username"] = $username;
// Redirect user to welcome page
header("location: asdasd.php");
} else{
// Display an error message if password is not valid
$password_err = "The password you entered was not valid.";
}
}
} else{
// Display an error message if username doesn't exist
$username_err = "No account found with that username.";
}
} else{
echo "Oops! Something went wrong. Please try again later.";
}
}
// Close statement
mysqli_stmt_close($stmt);
}
// Close connection
mysqli_close($link);
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Login</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.css">
<style type="text/css">
body{ font: 14px sans-serif; }
.wrapper{ width: 350px; padding: 20px; }
</style>
</head>
<body>
<div class="wrapper">
<h2>Login</h2>
<p>Please fill in your credentials to login.</p>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
<div class="form-group <?php echo (!empty($username_err)) ? 'has-error' : ''; ?>">
<label>Username</label>
<input type="text" name="username" class="form-control" value="<?php echo $username; ?>">
<span class="help-block"><?php echo $username_err; ?></span>
</div>
<div class="form-group <?php echo (!empty($password_err)) ? 'has-error' : ''; ?>">
<label>Password</label>
<input type="password" name="password" class="form-control">
<span class="help-block"><?php echo $password_err; ?></span>
</div>
<div class="form-group">
<input type="submit" class="btn btn-primary" value="Login">
</div>
<p>Don't have an account? <a href="register.php">Sign up now</a>.</p>
</form>
</div>
</body>
</html>这样，我就可以登录到由felvetel.php创建的帐户：
<?php
$teljes_nev = $_POST['teljes_nev'];
$kategoria = $_POST['kategoria'];
$felhasznalonev = $_POST['felhasznalonev'];
$idcske = "NULL";
header("Content-type: text/html; charset=utf-8");
if (!$teljes_nev) {
echo "Kérem írja be a diák nevét.";
exit;
}
if(!get_magic_quotes_gpc()) {
$teljes_nev = addslashes($teljes_nev);
$felhasznalonev = addslashes($felhasznalonev);
$jelszo_alap = rand(1000,9999);
$jelszo = password_hash($jelszo_alap, PASSWORD_DEFAULT);
}
@ $adatbazis = new mysqli('localhost', 'root', '', 'diakok');
$adatbazis->set_charset('utf8');
if (mysqli_connect_errno()) {
echo "Nem sikerült csatlakozni az adatbázishoz. (1-es hibakód)";
}
$lekerdezes = "INSERT INTO diakok VALUES
('".$felhasznalonev."', '".$teljes_nev."', '".$jelszo."', '".$kategoria."', '".$idcske."', '".$idcske."' , '".$idcske."' ,'".$idcske."', '".$idcske."')";
$talalat = $adatbazis->query($lekerdezes);
if ($talalat) {
echo $teljes_nev." nevű diák sikeresen felkerült az adatbázisba! (A diák jelszava ".$jelszo_alap.")";
} else {
echo "Hiba! A diákot nem sikerült hozzáadni az adatbázishoz.";
}
$adatbazis->close();
?>我的问题是它显示“您输入的密码无效”。当我写下正确的密码时。例如，我创建了一个名为admin.pelda的测试用户，其不带password_hash()的密码是2161。现在，当我尝试登录该帐户时，它会给出无效密码消息。有什么办法可以解决这个问题吗？(我的config.php没问题)
Stack Overflow用户
发布于 2019-02-22 05:28:10
问题是散列的密码非常长，并且在数据库中，密码的最大长度被设置为15。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/54811607
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