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?
答案 0 :(得分:0)
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' ";