尝试通过获取详细信息来让用户登录,但不起作用,请检查此
这是我的连接代码
<?php
$server = "localhost";
$username = "root";
$password = "";
$database = "atm";
$conn = new mysqli($server ,$username ,$password ,"atm");
if(!$conn =""){
echo "connection is established<br>";
}
?>
我的主要文件代码是
<?php
require_once "../configmain.php";
if(isset($_POST['login'])){
$username = $_POST["username"];
$password = $_POST["password"];
$adminqry = mysqli_query($conn,"select 'admin_id' from 'admin_details' where STATUS = '1' and username = '$username' and password = '$password';")or die (mysqli_error($con));
while($admin = mysqli_fetch_object($adminqry)){
$adminid = $admin -> admin_id;
}
session_start();
$_SESSION['adminid'] = $adminid;
$adminid = $_SESSION['adminid'];
echo $adminid;
if(!$adminid){
header("Location: ./admin/adminmain.php");
exit();
}
else{
echo "login error";
}
}
?>
检查我的连接是否正常工作,如果我的数据库正常工作,但它也正常工作,正在与 xampp 服务器一起工作,请帮助我
你的问题是这一行:
if(!$conn ="")
您没有检查
$connection
是否为空,您实际上是将其设置为空字符串。
尝试将您的支票更改为:
if(!$conn->errno)
这将检查连接对象上是否有任何非零错误号。
试试这个
第 1 步:用下面给定的代码替换您的连接文件
<?php
$server = "localhost";
$username = "root";
$password = "";
$database = "atm";
$conn = new mysqli($server ,$username ,$password ,$database);
?>
第 2 步:用下面的代码替换您的主文件代码
<?php
require_once "../configmain.php";
if(isset($_POST['login'])){
$username = $_POST["username"];
$password = $_POST["password"];
$adminqry = mysqli_query($conn,"select 'admin_id' from 'admin_details' where STATUS = '1' and username = '$username' and password = '$password';")or die (mysqli_error($conn));
if(mysqli_num_rows($adminqry) == 1){
$admin = mysqli_fetch_object($adminqry);
$adminid = $admin -> admin_id;
session_start();
$_SESSION['adminid'] = $adminid;
if(isset($_SESSION['adminid'])){
header("Location: ./admin/adminmain.php");
exit();
}else{
echo "Login error";
}
}else{
echo "User not found.";
}
}
?>
//Notes:
//(1) Check your connection file path is proper in require_once.
//(2) I consider you want to redirect header("Location: ./admin/adminmain.php"); here if login success.