如何在PHP程序中调试“警告：session_regenerate_id（）：无法重新生成会话ID”？

Question

I am making a messaging system, where the message id is session_id and it is on a session_regenerate id.

It's working fine but, when I changed my template, it keeps on giving errors like below :

" Warning: session_regenerate_id(): Cannot regenerate session id - headers already sent in C:\\xampp\\htdocs\\READS Website MAIN\\Admin\\admin-page\\admin\\message.php on line 24"

Below is my code:

<?php
include('header.php');
include('config.php');


if(isset($_POST['submit'])){

$to = $_POST['to'];
$subject = $_POST['subject'];
$message = $_POST['message'];
$id = session_id();
$sender = $_SESSION['id'];

date_default_timezone_set("ASIA/MANILA");
$date = date('m-d-Y h:i a');




$sql = "INSERT INTO `messages`(`session_id`, `sender`, `recipient`,   
`subject`, `content`, `date`, `stat`) 
VALUES ('$id','$sender','$to','$subject','$message','$date','unread')";
mysql_query($sql) or die(mysql_error());
session_regenerate_id();
if($sql){

echo "<script>alert('Message Sent')</script>";

}else {
echo "<script>alert('error')</script>";
}

}

?>
<div>
<ul class="breadcrumb">
<li>
<a href="#">Home</a>
</li>
<li>
<a href="#">User</a>
</li>
</ul>
</div>

<div class="box-content">


<form action='' method='POST'>

To: <select name='to' class="form-control" id="inputEmail3">

<?php $sql = "SELECT * FROM users ";
$result=mysql_query($sql); 
while ($row = mysql_fetch_array($result)){
echo "<option value='".$row['id']."'>".$row['Fname']."      
".$row['Lname']." (".$row['user_type'].")</option>";

}

?>
</select>


<div class='col-md-2 col-sm-2 col-xs-5' style='margin-left:150px;    
margin-top:-10px; width:500px;'>

<br>
Subject: <input type='textfield' name='subject' class="form-control"        
id="inputEmail3"><br><br>
Message: <textarea name='message'  class="form-control" id="inputEmail3"             
cols='30' rows='8'></textarea>
<div style = 'margin-left:420px; margin-top: 10px;'>
<input type='submit' name='submit'  value='Send'></div>
</td>




</body>
</html>
<?php include('footer.php');
?>

Answer 1

Make sure you are setting sessions or cookies before any content is sent. Sessions and cookies are set in the header of a response, and since you can't change a header after the content (body) has been sent, you are faced with this warning.

And after reading your code more thoroughly, I must remind you that this code should not be put into the real world, since you're not escaping any queries. Plus that you should really switch to MySQLi or PDO, since the mysql_* functions are deprecated since a few PHP-versions.