我需要为我的网站制作一个登录表单。而且我必须使用 MySQLi,因为 MySQL 会导致我的尝试失败。所以,这是索引.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$username="admin"和$password="admin"时,它会main_menu.php正常。但是当我尝试按照上述方式执行时,基于我的数据库,它不会main_menu.php。如何登录,使用数据库中的 ID 转到main_menu.php?
对不起,我已经检查过了,这是一个愚蠢的错误。在此代码段中:
$username = $_POST['userid'];
$pass = $_POST['password'];
$query1 = "select * from admintb where adID = '$username' and adPass = 'password' ";
$result1 = mysqli_query($con,$query1) or die;
修复此问题:
$query1 = "select * from admintb where adID = '$username' and adPass = '$pass' ";