PHP代码检查登录从数据库


php code for checking login from db

通过使用下面的脚本,我将用户注册到我的数据库

<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<title>Regsitration Successful</title>
<link rel="stylesheet" type="text/css" href="styles.css">
</head>
<body>
<div class="container">
<header>
<div id="logo"><a href="index.html"><img src="smu-logo.png"></a></div>
    <ul class="nav_menu">
        <li><a href="index.html">Home</a></li>
        <li><a href="register.html">Register</a></li>
        <li><a href="updates.html">Notices And Updates</a></li>
        <li><a href="reachus.html">Reach Us</a></li>
    </ul>
    <div class="clear"></div>
</header>
<h3>Success!</h3>

<?php
$f_name = $_POST['f_name'];
$m_name = $_POST['m_name'];
$l_name = $_POST['l_name'];
$reg_num = $_POST['reg_num'];
//$dept = $_POST['dept'];
$password = $_POST['password'];
$password2 = $_POST['password2'];
//$about = $_POST['about'];
//$etype = $_POST['etype'];
//connect code 
$conn = mysqli_connect("localhost","root","");
$db = mysqli_select_db($conn, "tnp");
$cmd = "insert into ug_login_details values   ('$reg_num','$password','$f_name','$m_name','$l_name')"; 
//for more columns add more after comma
//$cmd = "insert into ug_login_details values ('$reg_num','$password')";// adding user login credentials
if(mysqli_query($conn, $cmd))
{
   echo "Quick Registration Successful";
}
else
{
   echo "error";
}
echo "Dear, $l_name" ; //change this
?>

    <p>Kindly check for the next notice on our Notice and Updates page to confirm.</p>
    <p><a href="fullreg.html">Click here</a> to migrate to the complete registration form.</p>
</div>    
</body>
</html>

现在,我需要什么代码来检查用户登录时的凭据?如何比较数据库中的值?我需要创建一个新的页面吗?我已经有一个登录框在我的主页,索引。html。我是初学者,所以请应付一下。由于

你创建一个新页面是完全正确的。

你可以有一个索引页:

<form action="index.php" method="POST">
    Username:<input type="text" name="username"/>
    Password:<input type="text" name="password"/>
    <input type="submit" value="Login"/>
</form>

然后在PHP中:

  if($_POST)
    {
        $Username=$_POST["username"];
        $Password=$_POST["password"];
        $conn = mysqli_connect("localhost","root","");
        $db = mysqli_select_db($conn, "tnp");
        $cmd= "select * from ug_login_details where password='$Password' and username='$Username'";
        $Result=mysqli_query($conn, $cmd);
        $num=mysqli_num_rows($Result);
        if($num>0)
        {
            echo "You have successfully logged in";
            exit();
        }
        else
        {
            echo "Invalid username or password";
            exit();
        }
    }