hide Image/File Path Help

[uppalakishore]

First of All Thanks For Reading...

 

I Want to hide Image/File path, mainly i want to hide the "image path" there by Logged in users only see the images.

 

1) My Problem is When a user Logged in and opend his personal photos, the image path will be "http://localhost/photo_gallery/User_data/Personal_photos/username/kicku.jpg" .

 

2)But when i entered the path in the browser  without login the images keep displaying...

 

3)So i want to hide the image path for security purpose...

 

Any help, Appreciable...

 

 

[haku]

You cannot hide image paths. The browser needs to know the path to show the image, and once the browser has it, it's there for the user to see.

 

However, you can set up a system to serve your images from a php script, and then add security checks to that php script. So if the user has permission to see the image, the script will serve the image to the user, or if they don't have permissions, it will block them. The user can still see the image path (which is actually a path to a php script), but since the security check will happen if they try to access that path, it solves your problem.

 

Here is someting to get you started: http://www.thewebsqueeze.com/web-design-tutorials/how-to-output-images-using-php.html

[uppalakishore]

Thank u @ haku....

I will try....

[uppalakishore]

@ haku....

---

I tried and rename the directory as "Images" but i found it displaying the image path as it is and it will randomize the images in the directory when i refresh the page...

 

What i want is I want to hide the location/path of the image...

sum where i got some code i will try it...

 

---

 

<?php

 
  $imagedir = "Images/" ;

  $validprefixes = array (
    "http://localhost/photo_gallery/index.com",
    "localhost/photo_gallery/"
  ) ;

 
  $homepage = "index.php" ;

  function isreferrerokay ( $referrer, $validprefixes )
  {
    $validreferrer = 0 ;
    $authreferrer  = current( $validprefixes );
    while ($authreferrer) {
      if (eregi( "^https?://$authreferrer/", $referrer )) {
        $validreferrer = 1 ;
        break ;
      }
      $authreferrer = next( $validprefixes );
    }
    return $validreferrer ;
  }

  //----------------------- main program -----------------------

  $image = $_GET['image'] ;
  $referrer = getenv( "HTTP_REFERER" );

  if (isset($_GET['image'])) {

    if (empty($referrer) ||
      isreferrerokay( $referrer, $validprefixes )) {

      $imagepath = $imagedir . $image ;

      $imageinfo = getimagesize( $imagepath );
      if ($imageinfo[2] == 1) {
        $imagetype = "gif" ;
      }
      elseif ($imageinfo[2] == 2) {
        $imagetype = "jpeg" ;
      }
      elseif ($imageinfo[2] == 3) {
        $imagetype = "png" ;
      }
      else {
        header( "HTTP/1.0 404 Not Found" );
        exit ;
      }

      header( "Content-type: image/$imagetype" );
      @readfile( $imagepath );

    }
    else {

      if (isset($email)) {
        mail( $email, "Bandwidth Theft Alert",
           "WARNING:\n\n$referrer\ntried to access\n$image\n",
           "From: CHImageGuard <$email>" );
      }
      header( "HTTP/1.0 404 Not Found" );
    }
  }
  else {
    header( "Location: $homepage" );
  }

?>
 

Edited March 23, 2013 by uppalakishore

[uppalakishore]

The above code is working fine in case of a particular image like

freak_test.php?image=f3.jpg (The code is saved in a page "freak_test.php")

In case of,

<img src="freak_test.php?image=f3.jpg" width=200 height=300 />

It will not displaying the image

And when Right click on the image and click "View Image" it displays the following error

 

Deprecated: Function eregi() is deprecated in F:\xampp\htdocs\photo_gallery\User_data\User_details\freak_test.php

 

any Suggestions?

Edited April 6, 2013 by uppalakishore

[trq]

As the error says, eregi is deprecated.

 

There are details on the manual page about how to handle this.

[l0gic]

It's been awhile but I'm pretty sure you could also use .htaccess to only allow access to the directory from a certain referrer.

 

For example, allow access from ..yoursite/viewpic.php where the viewpic.php checks login status, and if all conditions are met redirects you to the image passed via GET/Query which you will be able to see because your referrer is viewpic.php if conditions are not met in viewpic.php, redirect to a 'you need to be logged in to view images' page/error.

 

So then if you had the image testpic01.jpg you would link to it such as ..href="path/to/viewpic.php?p=testpic01&f=jpg".. 'p' being the filename, 'f' being the format/extension.

 

I'm pretty sure you could also display images on a page using something similar to this method, and of course it could be used to 'protect' any file-type.

 

RewriteEngine On
RewriteCond %{HTTP_REFERER} !(www.)?example.com/viewpic.php
RewriteRule .* - [F]

 

Something like ^that anyway.

 

See Here: http://www.hongkiat.com/blog/smarter-way-to-prevent-image-hotlinking-with-htaccess/

Edited April 6, 2013 by l0gic

[uppalakishore]

Thank u @ trq and l0gic...
But i slightly confused...
 

---

I found a link http://michael.theirwinfamily.net/articles/csshtml/protecting-images-using-php-and-htaccess

 

I added the following .ht access files to Image directory and to main directory
 
.htaccess in main folder:

RewriteEngine on
RewriteCond %{HTTP_REFERER} ^$
RewriteCond %{SCRIPT_FILENAME} image\.php
RewriteRule (.*) image.php?onlyHappensFromHTACCESS=denied [QSA,L]

.htaccess in Images folder:

#Prevent directory listing
Options -Indexes
#Prevent images from being viewed
<Files *>
  deny from all
</Files>

image.php:

---

 

<?php
if (!isset($_GET['onlyHappensFromHTACCESS'])) {
   $_GET['f'] = "./Images/" . $_GET['f'];                            //my folder name="Images";
    $type = getFileType($_GET['f']);
    if (acceptableType($type)) {
        if (goodTiming()) {
            header("Content-type: $type");
         echo file_get_contents($_GET['f']);
            exit;
        }
    }
    header('HTTP/1.1 403 Forbidden');
    exit;
}
function getFileType($file) {
  if (function_exists("mime_content_type"))
    return mime_content_type($file);
  else if (function_exists("finfo_open")) {
    $finfo = finfo_open(FILEINFO_MIME_TYPE);
    $type = finfo_file($finfo, $file);
    finfo_close($finfo);
    return $type;
  }
  else {
    $types = array(
      'jpg' => 'image/jpeg', 'jpeg' => 'image/jpeg', 'png' => 'image/png',
      'gif' => 'image/gif', 'bmp' => 'image/bmp'
    );
    $ext = substr($file, strrpos($file, '.') + 1);
    if (key_exists($ext, $types)) return $types[$ext];
    return "unknown";
  }
}
function acceptableType($type) {
    $array = array("image/jpeg", "image/jpg", "image/png", "image/png", "image/gif");
    if (in_array($type, $array))
        return true;
    return false;
}
function goodTiming() {
    $n = time();
    session_start();
    if ($n - $_SESSION['lastcheck'] > 2 )
        return false;
    return true;
}
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<html>
<head>
  <meta http-equiv="Content-type" content="text/html; charset=utf-8" />
  <title>Image Denied</title>
  <style type="text/css" media="screen">
    body {
        background-color: #ccc;
        font-family: Helvetica, Arial;
    }
    #wrapper {
        margin: 30px auto;
        background-color: #ffffff;
        -moz-border-radius: 15px;
        -webkit-border-radius: 15px;
        border-radius: 15px;
        width: 800px;
        padding: 20px;
    }
  </style>
</head>
<div id="wrapper">
  <h3>Access Denied!</h3>
  You have tried to access an image, but due to security reasons, you cannot view the image.
  If you wish to use the image you requested, please contact me.
</div>
</html>

---

index1.php:

---

 

<?php session_start(); $_SESSION['lastcheck'] = time(); ?>
<html>
<head>
    <title>Page Title</title>
    <style type="text/css">
        .image {
            overflow: hidden;
            position: relative;
            float: left;
        }
        .image .cover, .image .cover img {
            position: absolute;
            top: 0px;
            left: 0px;
            width: 100%;
            height: 100%;
        }
    </style>
</head>
<body>
    <div class="image">
        <img src="image.php?f=f3.jpg" alt="Image" />
        <div class="cover"><img src="Transparent.gif" alt=""  /></div>
    </div>
</body>
</html>

---

I got the output as "Image"(Means alt will be loaded)...
when i right click on it,and select View Image "Transparent.gif" will be displaying normally.
 
I can't understant what is the problem here...
Edited April 8, 2013 by uppalakishore

[uppalakishore]

I want to display the image when i use the tag

 

<img src="image.php?f=f3.jpg" alt="Image" />

Any help Wil be Appreciable...