没有这样的文件或目录：我该如何解决这个问题？

Question

当我试图创建一个登录系统时，我收到了这个错误： Impossibile interrogare il database No such file or directoryPHP过程代码是这样的：<?php    session_start();include('db-conn-dashboard.php');$nomeutente = $_POST['nomeutente'];$password = $_POST['password'];$nomeutente = stripcslashes($nomeutente);$password = stripcslashes($password);$nomeutente = mysql_real_escape_string($nomeutente);$password = mysql_real_escape_string($password);$result = mysql_query("SELECT * FROM profili WHERE nomeutente = '$nomeutente' and password = '$password'")    or die("Impossibile interrogare il database ".mysql_error());$row = mysql_fetch_array($result);if($row['nomeutente'] == $nomeutente && $row['password'] == $password){    $_SESSION['login_user1'] = "autorizzato1";    $_SESSION['autorizzato'] = 1;    header('location: ../pages/home.php');} else {    echo '<script type="text/javascript">alert("Nome utente o password errati. Riprova.")    window.location= "../pages/login.php"</script>';}?>形式是这样的：<form role="form" action="../php/processo-accesso.php" method="POST">                            <fieldset>                                <div class="form-group">                                    <input class="form-control" placeholder="&#xf007;  Nome utente" name="nomeutente" type="text" style="font-family: fontawesome, arial;" onkeyup="this.value = this.value.toUpperCase();" required>                                </div>                                <div class="form-group">                                    <input class="form-control" placeholder="&#xf084;  Password" style="font-family: fontawesome, arial;" name="password" type="password" required>                                </div>                                <input type="submit" name="accedi" class="btn btn-primary" value="Accedi" style="width: 100%;">                            </fieldset>                        </form>

Answer 1

尝试这个

$sql= mysql_query("SELECT * FROM profili WHERE nomeutente = '$nomeutente' and password = '$password'")

    or die("Impossibile interrogare il database ".mysql_error());

$result=mysql_query($sql);

$row=mysql_fetch_array($result);