從具有偽造數據的數據庫中進行SELECT會產生大於零的計數

Question

即時通訊試圖創建一個登錄腳本，我覺得我幾乎在那兒，但是由於某種原因，當我通過輸入不正確的用戶名和密碼進行測試時，它仍然認為它從數據庫中得到了一些東西。

這是我的表單代碼：

        <form method="post" action="checklogin.php">
            <table>
                <tr>
                    <td><label style="color:#6b6a6b;" for="username">Username: </label></td>
                    <td><input class="textbox" type="text" name="username"></td>
                </tr>
                <tr>
                    <td>&nbsp;</td>
                    <td></td>
                </tr>
                <tr>
                    <td><label for="password">Password: </label></td>
                    <td><input class="textbox" type="password" name="password"></td>
                </tr>
                <tr>
                    <td>&nbsp;</td>
                    <td></td>
                </tr>
                <tr>
                    <td></td>
                    <td><input type="submit" name="login" id="login" value=""></td>
                </tr>
            </table>
        </form>

這是我的PHP代碼：

session_start();
$con=mysqli_connect("********", "*******", "******", "******");
$myroot = "";
$previousURL = parse_url($_SERVER['HTTP_REFERER'],PHP_URL_PATH);

if(isset($_POST['login'])){
    if(isset($_POST['username']) && isset($_POST['password']) && $_POST['username'] != "" && $_POST['password'] != ""){
        if (strlen($_POST['username']) > 20 || strlen($_POST['username']) < 4)
        {
            $error = 'Incorrect Length for Username or Password';
        }
        /*** check the password is the correct length ***/
        elseif (strlen($_POST['password']) > 20 || strlen($_POST['password']) < 4)
        {
            $error = 'Incorrect Length for Username or Password';
        }
        /*** check the username has only alpha numeric characters ***/
        elseif (ctype_alnum($_POST['username']) != true)
        {
            /*** if there is no match ***/
            $error = "Username must be alpha numeric";
        }
        /*** check the password has only alpha numeric characters ***/
        elseif (ctype_alnum($_POST['password']) != true)
        {
            /*** if there is no match ***/
            $error = "Password must be alpha numeric";
        }

        $username = $_POST['username'];
        $password = $_POST['password'];
        $username = stripslashes($username);
        $password = stripslashes($password);
        $username = mysqli_real_escape_string($con,$username);
        $password = mysqli_real_escape_string($con,$password);
        $result=mysqli_query($con,"SELECT * FROM RAE_customers WHERE username='" . $username . "' AND password='" . $password . "'")  or die(mysqli_error($con));

        $count=count($result);

        if($count==1){

            // Register $username
            $_SESSION['username'] = $username;
            var_dump($_SESSION['username']);
            echo "<br>";
            echo $count;
            /*if($previousURL == "/checkout.php"){
                header('Location: details.php');
            }
            elseif($previousURL == "/login.php"){
                header('location: index.php');
            }*/

        }
        else {
            $error = "Wrong Username or Password.";
        }
        if(isset($error)){
            echo $error;
        }
        else{
            echo "success";
        }
    }
    else{
        $error = "Please enter a Username and Password.";
    }

}
else{
    header('location:index.php');
}

當我輸入虛擬的用戶名和密碼時，它會將其視為正確。

我得到這些結果

字符串（6）“ dffddf” 1成功

有人看到我的代碼有問題嗎？

Answer 1

$result是一個mysqli結果對象，因此count($result)將始終為1 （除非查詢失敗）。

嘗試使用：

$result->num_rows;

代替。

Answer 2

不要使用count count僅適用於數組，而您得到的是mysqli結果，而不是mysqli_num_rows，因此

$count=count($result);

變成

$count=mysqli_num_rows($result);