I put the little code on the header of php file to blocking proxies:
if (isset($_SERVER['HTTP_X_FORWARDED_FOR'])
|| ($_SERVER['HTTP_USER_AGENT']=='')
|| ($_SERVER['HTTP_VIA']!='')) {
die("Don't use proxies, please.");
}
In member.php I put the above code and its work very well and when someone request example.com/member.php with http proxy this code blocks them, but when they request example.com/member.php?action=login this code can't block them! What am I going to do? Thanks in Advance.
A blank user agent doesn't imply they're using a proxy... some people just don't like broadcasting what browser/OS they're using... As for why the login script doesn't get blocked why not look at what's in $_SERVER at each stage. A simple var_dump($_SERVER) will show everything and tell you why what you're trying to match isn't being matched.
Related
I have a flask app that redirects to a php site on an Apache server. I don't have much experience with php. The flask app is at mysite.xyz and the php site is at mysite.xyz/page1. I would like to deny access to anyone going to mysite.xyz/page1 unless they are coming from mysite.xyz. Something along the lines of:
if not coming from 'mysite.xyz':
redirect to 'mysite.xyz'
else:
pass
Except at the beginning of my php and in php.
Look at http://php.net/manual/en/reserved.variables.server.php and check the 'HTTP_REFERER' part.
Example Code:
<?php
if($_SERVER['HTTP_REFERER'] == "mysite.xyz"){
// ok, continue to load the page...
}else{
die("not allowed");
}
However, you can SPOOF the referrer to make it seem to be coming from mysite.xyz (but it is not the case) so it will never be secure.
Is there an option to provide an authentication system to your users to be able to handle this functionality without security flaws?
I've been trying to get the URL (including GET parameters) of a site that is displaying my image. This is because I want to extract one parameter of the URL.
A friend told me that she knew someone that could achieve this, but I don't know if he was doing it with an image. Also I don't think I can do it with a link because when going to external sites it will appear a warning page saying that you're being redirected outside, so if I put a link to my page and someone clicks, I will get the referrer URL of redirection warning page. I can't assure if my friend was telling the truth about this, but it's very likely that it was true.
All I could get with the image was the IP and other things of the HTTP header, but the referrer part is empty and I thought that the referrer contained the full URL I'm talking about.
This is what I have tried.
First the img tag in the other site in BBCode:
[img]http://______.com/get_image.php?i=myimage[/img]
And in my site this script in PHP, although any language that does the work would be good for me:
<?php
// Get name of image to be displayed (non-sanitized here for simplicity)
$filename = $_GET["i"];
// Here I want to get the site where image is being viewed
if (!empty($_SERVER['HTTP_REFERER'])) {
$visitor_url = $_SERVER['HTTP_REFERER'];
} else {
$visitor_url = "none";
}
// And write the referrer to a file just to test if it works
$fp = fopen('referer.txt', 'w');
fwrite($fp, $visitor_url);
fclose($fp);
// Eventually display the image
header('Content-Type: image/png');
readfile($filename . '.png');
?>
So my questions are:
Is it possible to get full URL of a site that is displaying my image?
If not, is there any other method to get the full URL?
Thank you in advance.
Note: I don't have any permision in the other site where I'm posting the image, I'm just an user there. Please tell me if I'm missing something or I have to ask this in another way, I'm new to StackOverflow.
Try REMOTE_HOST instead of HTTP_REFERER:
// Here I want to get the site where image is being viewed
if (!empty($_SERVER['REMOTE_HOST'])) {
$visitor_url = $_SERVER['REMOTE_HOST'];
} else {
$visitor_url = "none";
}
The web server where you are serving the image will need to be configured properly. If using Apache, this is with HostNameLookups On.
See http://php.net/manual/en/reserved.variables.server.php
Normally browsers are sending full referer with all URL components including query parameters - $_GET params. If they don't then there is no other way to achieve that URL while passing throught an image content.
Sometimes sending referer may be blocked, for eg. in some batch URL processing using some crawler like program/script or on some proxies.
In PHP receiving referer is done by $_SERVER['HTTP_REFERER'] because it's normally just http header from request and it's the only $_SERVER array key with referer info.
You added the .htaccess tag so I think you're using the Apache web server. If you'd like to prevent the issue entirely, you can disable hotlinking entirely by going one layer lower. Instead of managing in PHP, you can configure the web server to not serve content to domains other than the one you are hosting.
Check out the guide for more details.
I fixed this problem by switching my site (where image is hosted) to HTTPS. The code in my question was doing its job correctly.
It looks that HTTP_REFERER was blank because of it coming from an HTTPS site and my site being HTTP it would always send it blank. I was aware that it could be a problem, but didn't make much sense for me because HTTP_REFERER was also blank when coming from another HTTP site (which I think it's not normal) so I thought the error was in another place.
Usually HTTP_REFERER is sent when it comes from and goes to:
from HTTP to HTTP
from HTTPS to HTTPS
from HTTP to HTTPS
But it's not sent when it comes from and goes to:
from HTTPS to HTTP
And in my case, I don't know why, it wasn't being sent from HTTP to HTTP which was confusing me.
I created a PHP page that is allowed only for one person, That's OK but when the non allowed user write in address bar view-source:tstwebsite.com/test/page.php it shows the user the source of the page can I block the source code from the user? this is my code that allow this page only for one user.
$aiwab = mysql_query("SELECT * FROM `table`");
while($aiwa = mysql_fetch_array($aiwab)){
$alo = $aiwa['allowed'];
if ($alo == 2 ){
}else{
echo "<script>javascript:history.go(-1)</script>";
}
}
So how Can I block the user from viewing the source code?
Using javascript to "block" a user is, frankly, stupid. javascript can be disabled/ignored/bypassed, exactly as you've seen with your view-source "hack". "Security" can never be established if you're relying on the CLIENT to cooperate.
Use proper server-side authentication, e.g. even HTTP basic authentication, to protect the script.
If you just want to quickly change this from client-side authentication to server-side, then you could make the following change:
if ($alo == 2) {
}
else {
// Redirect them by sending a HTTP Location header
header("Location: www.yourdomain.com/path/to/another/page");
}
Note that the above solution will only work if header() is called before any output is sent to the browser (HTTP headers have to be sent before the body of the message begins).
This functions very similarly to your current solution, with the difference that the redirect is caused by code on the server rather than in the browser. view-source lets someone get around your authentication as it allows them to load the page in their browser, without running the client side code.
This is just a quick fix that should help illustrate the difference between client side and server side authentication. If this is for anything beyond just messing about and learning a little code, then you should really be learning more about security. Also note that the mysql functions you're using are currently deprecated and you should instead be using mysqli, or pdo
You will also want to read up on the uses of client-side versus server-side code, just to get a grasp of what to use for what tasks and why.
a website has used a "301 permanent redirect" to my site is there a way i can set code that detects this and displays a page when my website is accessed through this?
Does anyone have any idea about this?
You can get only a referer. I think you will not be able to get the http status code on server which the client gets during last request.
So my answer is NO, you cannot get the 301 status code on your server.
But you can do a little of needed magic with referer variable.
e.g. in PHP you can read this:
$_SERVER['HTTP_REFERER'];
Not much you can do. If you were doing the 301, you could set the referrer to the querystring. But since you're not, you can only grab what the request has given you.
You can try using PHP's $_SERVER['HTTP_REFERER'] to track the source URL from where your visitor comes from. I think it's a bit dodgy though and might not yield the same result in all browsers. Even PHP's documentation says 'it cannot really be trusted'.
Why do you have to use .htaccess for the redirect? You could do something like this:
Site A's index.php:
header("Location: http://siteb.com/?ref=".urlencode('http://sitea.com');
Site B's index.php:
if(isset($_GET['ref']))
{
if($_GET['ref']=='http://sitea.com')
{
// Do something
}
}
Edit:
If you can't edit Site A's code or server settings, try using:
if($_SERVER['HTTP_REFERER']=='http://sitea.com')
{
// Do something
}
one of my php page returns data like this:
<?php
//...
echo "json string";
?>
but someone else use file_get_contents() to get my data and use in other website.
can anybody tell me what can i do to prevent such thing happen.
i consider if i can get the request's domain name to echo something else.but i dont know
the function to get request's domain name.and if the request is sent by a server,that
will be unhelpful. My English is poor, to express doubts, please bear with.
you can also use sessions. if somewhere in your application, before the user gets the json data, you start a session, then in this page where you are outputting json data, you can check for the session variable. this way only users that have passed the session generator page, can view your output.
suppose you have page A.php that generates the session. use this code before outputting anything in this page.
session_start();
$_SESSION['approvedForJson'] = true;
then in your page where you are outputting json data, before outputting anything, call session_start() again. the beginning of your PHP code is a good place to call it.
then before outputting the json data, check if the session variable for approved users exists, or not.
if ( isset($_SESSION['approvedForJson']) && $_SESSION['approvedForJson'] ) {
echo "json data";
} else {
// bad request
}
You can use $_SERVER['REMOTE_ADDR'] to get the address of the client address. You can also check $_SERVER['HTTP_REFERER'] and block external requests that way, but it's less reliable. There's probably a few other techniques involving $_SERVER that you can try.
Your fighting an uphill battle here. I am assuming your serverside process that responds in json is being consumed via javascript in your users browsers... so there is no easy way to encrypt it. You might try some of the techniques used to prevent xspf (see http://en.wikipedia.org/wiki/Cross-site_request_forgery ). If you developed the client to pass along some session token that is uniq per client you could reduce some of the problem. But, chances are whoever is stealing your data is gonna figure out whatever mechanism you put in place ... assuming this is some sort of ajax type thing. If its a server-server thing then as sli mentions, setting up some restrictions based on the remote ip would help, plus setting up some sort of API authentication tokens would help even more (see oauth for some pointers)
You could also using .htaccess with apache block every external request to the page if it get's called internally or block every request that is not from your domain:
Google search thingie
EDIT
You could also use some php file which includes the file which can not be read. So for example you have file.php:
<?php
$allowedFiles[] = 'somefile.php';
$allowedFiles[] = 'someotherFile.php';
$allowedFiles[] = 'jsonReturnFile.php';
if(in_array($_GET['file'], $allowedFiles)){
include( "include/".$_GET['file'] );
}
?>
Then you can allow file_ get _contents() on that file and write a rewriteRule in your .htacces to disallow any request to the include/ folder.
RewriteRule include* - [F,NC]
That will return a 403 forbidden error for a request to that directory or any file in the directory.
Then you can do you JSON request to something like: file.php?file=jsonReturnFile.php&someothherParamReadByJsonFile=1
And when someone tries to get the file contents for the JSON file they will get the forbidden error, and getting the file contents for the include.php won't return anything usefull.