Question

我需要为我的网站制作一个登录表单。我必须使用MySQLi，因为MySQL会在我的尝试中导致decaprated。所以，这是index.php代码：

<?php
session_start();ob_start();

$con=mysqli_connect("localhost","root","","oos");
if (mysqli_connect_errno()) echo "Failed to connect to MySQL: " . mysqli_connect_error();

    if(isset($_POST['signin']))
    {
        $username = $_POST['userid'];
        $pass = $_POST['password'];

        $query1 = "select * from admintb where adID = '$username' and adPass = 'password' ";
        $result1 = mysqli_query($con,$query1) or die;

        $co=0;

        while($row=mysqli_fetch_assoc($result1)) $co++;
        if($co==1)
        {
            $_SESSION['a']=$username;
            header("Location: main_menu.php");
        }
    }   ?>

问题是，当我使用$ username =“admin”和$ password =“admin”时，它会转到main_menu.php。但是，当我尝试按上述方式执行操作时，基于我的数据库，它不会转到main_menu.php。如何登录，使用我的数据库中的ID转到main_menu.php？

Answer 1

抱歉，我已经检查过，这是一个愚蠢的错误。在这个片段内：

 $username = $_POST['userid'];
    $pass = $_POST['password'];

    $query1 = "select * from admintb where adID = '$username' and adPass = 'password' ";
    $result1 = mysqli_query($con,$query1) or die;

解决此问题：

$query1 = "select * from admintb where adID = '$username' and adPass = '$pass' ";

无法使用我的数据库注册

1 个答案: