网站会话的 PHP isset 函数无法正常工作答案

【问题标题】：PHP isset function for website session not working correctly网站会话的 PHP isset 函数无法正常工作
【发布时间】：2020-04-15 21:41:59
【问题描述】：

所以我正在尝试为网站实现登录，我希望它更改顶部的菜单栏。但是我在使用 php 会话时没有得到想要的结果

我在开头使用 session_start

<?php
session_start();
?>

然后为了改变我使用的菜单栏

<?php
    if (!isset($_SESSION['username'])){
 ?>
<ul>
    <li><a href="Index.php">Home</a></li>
    <li><a href="About.php">About</a></li>
    <li><a onclick="document.getElementById('log').style.display='block'" style="width:auto;">Log In</a> </li>
</ul>
<?php
    }else if (isset($_SESSION['username'])){
?>
<ul>
    <li><a href="Index.php">Home</a></li>
    <li><a href="About.php">About</a></li>
    <li><a href="#logout">PLEASE LOG OUT</a></li>
</ul>
<?php
    }
?>

我的模式框和日志脚本如下

<div id="log" class="modal">
    <form class="modal-content animate" action="logindata.php" method="post">
        <div class="container">
            <label for="uname"><b>Username</b></label>
            <input type="text" placeholder="Enter Username" name="usrname" required>
            <label for="psw"><b>Password</b></label>
            <input type="password" placeholder="Enter Password" name="psw" required>
            <button type="submit">Login</button>
        </div>
        <div class="container">
            <button type="button" onclick="document.getElementById('id01').style.display='none'" class="cancelbtn">Cancel</button>
        </div>
    </form>
</div>
<script>
    // Get the modal
    var modal = document.getElementById('log');
    // When the user clicks anywhere outside of the modal, close it
    window.onclick = function(event) {
        if (event.target == modal) {
            modal.style.display = "none";
        }
    }
</script>

我的 logindata.php 包含以下内容

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "phpmysql";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
$user1 = $email1 = $pass1 = "";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
  $user1 = test_input($_POST["usrname"]);
  $pass1 = test_input($_POST["psw"]);
}  
function test_input($data) {
  $data = trim($data);
  $data = stripslashes($data);
  $data = htmlspecialchars($data);
  return $data;
}
$sql = "SELECT username, password, email FROM users";
$result = $conn->query($sql);
$row = $result->fetch_array();
if ($row["username"]==$user1 && $row["password"]==$pass1) {    
session_start();  
$_SESSION["username"] = $row["username"];
//$_SESSION["email"] = $row["email"];
header("Location: Main_login_authentication.php"); 
} else {
         header("Location: Denied.php"); 
}
$conn->close();
?>

我知道通过明文发送密码不是一个好策略。

【问题讨论】：

不要使用test_input()。这是一个无用的功能，会损坏您的数据。
切勿以明文形式或使用 MD5/SHA1 存储密码！ 仅存储使用 PHP 的 password_hash() 创建的密码哈希，然后您可以使用 password_verify() 进行验证。看看这个帖子：How to use password_hash 并了解更多关于bcrypt & password hashing in PHP

标签： php html authentication session logout

【解决方案1】：

session_start 需要位于代码的第一行...我看到您在另一页上有这个，但是这个仍然是错误的；）否则，在 isset 之前执行： print_r($_SESSION);

<?php
session_start(); 
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "phpmysql";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
$user1 = $email1 = $pass1 = "";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
  $user1 = test_input($_POST["usrname"]);
  $pass1 = test_input($_POST["psw"]);
}  
function test_input($data) {
  $data = trim($data);
  $data = stripslashes($data);
  $data = htmlspecialchars($data);
  return $data;
}
$sql = "SELECT username, password, email FROM users";
$result = $conn->query($sql);
$row = $result->fetch_array();
if ($row["username"]==$user1 && $row["password"]==$pass1) {     
$_SESSION["username"] = $row["username"];
//$_SESSION["email"] = $row["email"];
header("Location: Main_login_authentication.php"); 
} else {
         header("Location: Denied.php"); 
}
$conn->close();
?>

【讨论】：