Question

我设计了一个登录页面，我希望将用户名和密码与来自数据库的已保存数据相匹配，并打开仪表板页面，这是我的HTML代码。

> <form class="m-t" role="form" method="post" action="dashboard_4.html">
>                 <div class="form-group">
>                     <input type="email" class="form-control" placeholder="Username" class="form-control" required=""
> name="logemail">
>                 </div>
>                 <div class="form-group">
>                     <input type="password" class="form-control" placeholder="Password" class="form-control" required=""
> name="logpass">
>                 </div>
>                 <button type="submit" class="btn btn-primary block full-width m-b" id="login" name ="submit">Login</button>
> 
>                 <a href="login.html#"><small>Forgot password?</small></a>
>                 <p class="text-muted text-center"><small>Do not have an account?</small></p>
>                 <a class="btn btn-sm btn-white btn-block" href="register.html">Create an account</a>
>             </form>

这是我的PHP代码：

<?php  
$servername = "localhost";
$username = "sehnoqta_userbmc";
$password = "u?gQ=uS%t;a?";
$dbname = "sehnoqta_bmc";

if(isset($_POST['submit']))
{
$username = $_POST['logemail'];
$password = $_POST['logpass'];
$con=mysqli_connect("localhost","sehnoqta_userbmc","?gQ=uS%t;a?","rsehnoqta_bmc");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
$qz = "SELECT * FROM regis where email1='".$username."' and password3='".$password."'" ; 
$qz = str_replace("\'","",$qz); 
$result = mysqli_query($con,$qz);
$row = mysqli_num_rows($result);
if($row == 1)
  {
  header("location:new_page.php");
exit();
  }
mysqli_close($con);
}
?>

但它打开页面，即使我输入的用户名和密码不正确，我的代码或任何问题是否有任何问题......

Answer 1

您正在为连接使用不同的变量。使用相同的用户名和密码，否则您将被拒绝访问。我刚刚删除了你的CSS并简化了一些事情。（注意：这是提到的代码提到的问题的解决方案。它不是数据公开和安全原因的解决方案）

文件：login.php

<html>
<head>
<meta charset="utf-8">
<title>LogIn.php</title>
</head>
<body align="center">

 <form name="form" method="post" action="dashboard_4.php">
    <div >
        <input type="email" placeholder="Username" name="logemail">
    </div>

    <div >
        <input type="password" placeholder="Password" name="logpass">
    </div>

    <button type="submit" id="login" name ="submit">Login</button>

    <a ><small>Forgot password?</small></a>
    <p ><small>Do not have an account?</small></p>
    <a >Create an account</a>
</form>
</body>
</html>

文件dashboard_4.php

<?php
    $servername = "localhost";
    $username = "sehnoqta_userbmc";
    $password = "u?gQ=uS%t;a?";
    $dbname = "sehnoqta_bmc";

    $con = mysqli_connect("localhost","sehnoqta_userbmc","u?gQ=uS%t;a?","sehnoqta_bmc");
    // Check connection
    if (mysqli_connect_errno()){
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    if(isset($_POST['submit'])){
        $username = $_POST['logemail'];
        $password = $_POST['logpass'];

        $qz = "SELECT * FROM regis WHERE email1='$username' and password3='$password' ";
        //echo $qz."<br/>";

        $result = mysqli_query($con,$qz);

        //$temp=mysqli_fetch_assoc($result);
        //echo $temp['email1']." ".$temp['password3'];;

        $row = mysqli_num_rows($result);
        if($row == 1){
            header("location:new_page.php");
            exit();
        }//else {echo "no record combination";}
        mysqli_close($con);
    }
?>

Answer 2

为什么不尝试：

<?php  
$servername = "localhost";
$username = "sehnoqta_userbmc";
$password = "u?gQ=uS%t;a?";
$dbname = "sehnoqta_bmc";

if(isset($_POST['submit']))
{
$username = htmlspecialchars($_POST['logemail']);
$password = htmlspecialchars ($_POST['logpass']);

$con=mysqli_connect("localhost","sehnoqta_userbmc","?gQ=uS%t;a?","rsehnoqta_bmc");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$username = mysqli_real_escape_string($con , $username);
$password = mysqli_real_escape_string($con , $password);
$qz = "SELECT * FROM `regis` where `email1` = '$username' and `password3` ='$password'" ; 

$result = mysqli_query($con,$qz);
$row = mysqli_num_rows($result);
if($row == 1)
  {
  header("location:new_page.php");
exit();
  }
mysqli_close($con);
}
?>

匹配用户名和密码与mysql和PHP无法正常工作

2 个答案: