How to show image hidding its direct link in php? - php

Question

How can I create a php code that will process an ID for example and return corresponding image without providing client with actual image hotlink.
As implemented on link below the image is displayed in html page, but the hot link is hidden from client. Even opening the image in new window same link is shown not the original link to image.
This is definitely not implemented in .htaccess file as for each image URL with id corresponding image is rendered and is not a redirect to a single page.
http://www.imagesup.net/?di=15140291437113

Answer 1

a basic way could be something like this
<?php
// query to database mabye?
$myPath = getImagePath($_GET['id']);
// what kind of image file is that?
$contentType = getContentType($myPath);
// read file content
$img = file_get_contents($myPath);
// here you tell the browser what kind of image is that e.g. jpeg,png,gif...
header("Content-Type: $contentType");
echo $img;
?>
you need to define getImagePath and getContentType functions according to your needs.

Answer 2

Whenever a surfer enters one of my websites, I always assign a session to him. The session holds a couple infos, for example: agent, IP, language, date, … etc., and gets passed along via cookie or via GET (as parameter to each one of my pages).
Since I deal with a lot of image content, I started databasing my collection. Which basically means that for administration and clustering purposes, I am saving all my images to a SQL database which is multi-homed and spread accross several servers. One could argue if that is a smart thing to do, but we can argue that on another day and in another article.
I wrote a little script which is used throughout my site:
<img src="http://example.com/display.php?id=34" border="0" alt="" />
With an ever changing ID of course. That’s the part referencing my images in the database.
The following is the code from the script which I use to retrieve the image from the database:
<?php
$connection=#mysql_connect(...);
#mysql_select_db(...);
$query="SELECT mime, file FROM images
WHERE id=".$_GET["id"];
$rawdb=#mysql_query ($query,$connection);
if($rawdb AND #mysql_num_rows($rawd-->0){
$array=# mysql_fetch_array($result);
if (!empty($array["fileContents"])){
// Output the MIME header
header("Content-Type: ".$array["mime"]}");
// Output the image
echo $array["file"];
}else{
// something else...
}
#mysql_free_result($rawdb);
}else{
// something else...
}
#mysql_close($connection);
?>
Since I already have a session for each user that comes to my website, I just added the following:
<img src="http://example.com/display.php?id=34&sid=383829" border="0" alt="" />
And implement a small session checkup in the script itself:
<!--
session_start();
if($_SESSION["is_known"]){
// do database calls
}else{
header("Location:http://mydomain.tld/dontsteal.html");
}
-->
The main advantage to my method is, that the session is entirely server side. A user can not rid himself off it, or fake information. Since I have a timeout and save all the necessary info (IP!) to validate against, it looks pretty perfect to me and fit my needs.
One of the setbacks here are resources and performance. But since I am not forcing you, you may test and evaluate. Hope that helps!

Answer 3

Create a php script which you use as src in the img tag.
In that script get the data from the image with file_get_contents. Then send the header with the right mime type. For example header('Content-type: image/jpeg'); Then output the data.