PHP Mysql多用户登录表单 - PHP Mysql Multiple Users Login Form

PHP Mysql Multiple Users Login Form

本文关键字：表单登录多用户 Mysql PHP | 更新日期: 2023-09-26

如何检测用户是admin还是customer?下面显示的代码有一个if else语句，这被认为是错误的，因为我不知道它的代码是什么，你能帮我吗?我想检查$confirmation变量。如果包含1，则用户将进入管理页面。如果没有，则转到客户页面。我该怎么做呢?顺便说一句，if是管理员，然后第一个else是if else语句中的客户。

PHP代码:

<?php
session_start();
include("dbcon.php");
$username = $_POST['username'];
$password = $_POST['password'];
$qry="SELECT * FROM admininfo WHERE username='$username' AND password='$password'";
$result=mysql_query($qry);
    if($result) {
    if(mysql_num_rows($result) == 1) {
        session_regenerate_id();
        $user = mysql_fetch_assoc($result);
        //$_SESSION['fname'] = $user['fname'];
        //$_SESSION['lname'] = $user['lname'];
        //$_SESSION['username'] = $user['username'];
        //$_SESSION['password'] = $user['password'];
        $_SESSION['confirmation'] = $user['confirmation'];
        session_write_close();
        header("Location:menu.php");
        exit();
    }else {
        header("Location:menu1.php");
        exit();
    }
}else {
    die("Query failed");
}
?>

您可以做的是为admininfo表添加另一个字段，并在这里声明用户是否为admin。例如，在admininfo中有状态字段，您的查询将是这样的(这只是一个示例):

$yourconnection=mysqli_connect("YourHost","YourUsername","YourPassword","NameOfYourDatabase");
if(mysqli_connect_errno()){
echo "Error".mysqli_connect_error();
}
$username = mysqli_real_escape_string($yourconnection,$_POST['username']); /* prevents a bit of SQL injection */
$password = mysqli_real_escape_string($yourconnection,$_POST['password']); /* prevents a bit of SQL injection */
$qry=mysqli_query($yourconnection,"SELECT * FROM admininfo WHERE username='$username' AND password='$password' AND status='customer'");
$qryadmin=mysqli_query($yourconnection,"SELECT * FROM admininfo WHERE username='$username' AND password='$password' AND status='administrator'");
if(mysqli_num_rows($qry)==1){
header("LOCATION:CustomerPage.php");
}
else if(mysqli_num_rows($qryadmin)==1){
header("LOCATION:AdminUserPage.php");
}
else {
echo "Error! No user found.";
}

并尝试使用MySQLi代替它的前身MySQL