网站会话的PHP isset函数无法正常工作

Question

所以我试图实现一个网站的登录，我希望它能更改顶部的菜单栏。然而，当使用php会话时，我没有得到所需的结果

我在开始时使用session_start

<?php
session_start();
?>

然后为了更改菜单栏，我使用

<?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所示

<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

<?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。

Answer 1

session_start需要位于代码的第一行。。。我看到你在另一页上有这个，但这个仍然是错误的；(否则，在isset执行以下操作之前：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();
?>