Trying to see what actions can be performed with a PHP script that is being called via an image src like so:
<img src="http://example.com/script.php" />
Now, I have tried to include the PHP header() function in script.php:
<?php
header("Location: http://example.com");
I have also tried to echo an image url expecting the img to display it, which it didn't:
<?php
echo 'http://example.com/image.png';
Are there any ways of doing such things with a PHP script that is being called in the img src attribute?
Are there any ways of doing such things with a PHP script that is being called in the img src attribute?
No. A resource that is used as a src for an img tag needs to output image data, nothing else.
There are some exceptions, eg. a header("location: ....") redirect, but the redirect needs to point to another valid image resource, not a web site as you show in your example.
Check out the readfile() as a way to output your image file from your script.php
readfile($file);
Read more about it here in the manual:
http://php.net/manual/en/function.readfile.php
where Example #1 gives an idea of how to set up the headers.
The manual also states that:
readfile() will not present any memory issues, even when sending large
files, on its own.
and
A URL can be used as a filename with this function
ps: This was the way Wordpress Multisite used to open user uploaded (e.g. images) files.
Your script.php should return the output of an image with the correct headers. For instance:
<img src="/html/img/script.php" />
// Script.php
$file = "tiger.jpeg";
$type = "image/jpeg";
header("Content-Type: $type");
header("Content-Length: " . filesize($file));
readfile($file);
You should keep in mind that the src tag should directly point to an image file. However, it is possible to use PHP to create an image, for exmaple by using the GD library:
http://php.net/manual/en/book.image.php
So using:
<img src="http://example.com/script.php" />
is possible, as long as script.php really outputs an image file, for example by using the example as described here:
http://www.php.net/manual/en/image.examples-png.php
I used this kind of processing in the past to overlay texts on JPG images for a broker website (e.g. new, sold, for rent, etc.).
Are there any ways of doing such things with a PHP script that is being called in the img src attribute?
Yes, but the PHP Script has to output image data only, as stated in various other answers.
With that being said, just read the image and output it to the stream with readfile
header('Content-Type: image/png');
readfile($file);
exit();
I know I might be a couple years late to really help you, but the accepted answer just isn't true (anymore).
Related
I am trying to display img_a.png or img_b.png based on the access level of the user (e.g. signed in or not). Of course the content (in this case img_a and img_b) should not be available to the general public.
I tried a few solutions, none of them helps me and that's why I look for help here. What I tried so far is:
Based on the user checks I tried adding the resources folder to "open_basedir" which is a lame option but was looking the easiest solution. All ended up by raising a warning that resource is not in the basedir folder through it's obviously there.
Attempted to put the resources in the public folder and restrict them via .htaccess. But in this case they got restricted not only for the unwanted audience but for everyone.
Final and closest attempt was to put back the images outside the webroot and write a class that validates the access and serves the image like:
class Image {
...
public function getImage() {
...
header('Content-Type: '.$this->type);
readfile($this->item);
which afterwards is displayed in the initial script through:
echo "<img src=".$image->getImage($file).">";
The problem above is that the headers were already sent so I could either stream the image or the html output of the .php page but not both. Do I have a way out there?
Create a script that checks whatever user attribute you want, determines what image to serve, and read/send that image. Use the script URL as the <img src=... attribute, ie
<img src='/scripts/user_image.php'>
Something like this will work for PNG images. Similar functions exist for GIF, JPG, etc.
<?php
// do stuff here to determine file name of image to send
if($_SESSION['userlevel']=="ADMIN"){
$imageFilename="admin_image.png";
}
// Create Image From Existing File
$image = imagecreatefrompng($imageFilename);
//Set the Content Type
header('Content-type: image/png');
// Send Image to Browser
imagepng($image);
// Clear Memory
imagedestroy($image);
exit;
?>
OK, per your comment, I think you are referencing things wrong.
My working script is exact as above, only the if-then is commented out. I'm just assigning the filename to the variable. I've named the script user_image.php.
A simple index.html file to reference the image -
<html>
<head><title>test</title></head>
<body>
Hello World!<br />
<img src="user_image.php">
</body>
</html>
And it Just Works. See it in action at https://itsjustcrap.com/img or grab source at https://itsjustcrap.com/img/src.zip
I'll leave it up for a few days, but other than a png image and a commented out if-then, the source is exact same as what is in this post.
Actually I think I found a solution to the headers confict I had.
Displaying the image as data like:
<img src="data:image/png;base64, <?=base64_encode(file_get_contents($file))?>" alt="img">
instead of feeding it inside the source attribute with readfile() php script like:
<img src="image.php?<?=$file?>" alt="my_image">
where the last one was giving me a headers conflict when inside html page.
Discovered the solution over here How to display Base64 images in HTML?
If I use php file as source to image, where:
$file = $_GET["file"];
$file_get = get_file_contents("from/".$file);
$fopen = fopen("to/".$file,"w+");
fwrite($fopen, $file_get);
fclose($fopen);
header("Location:to/".$file);
And if I use many images of that kind on one page, like:
<img src="image.php/?file=img.jpg>
<img src="image.php/?file=img2.jpg>
<img src="image.php/?file=img3.jpg>
...
I found that code in image.php doesn't run asynchronously. Images are downloaded one by one. How can I avoid it?
I see there some problems in your code. The first is that you have a big security whole when you use the $_GET input directly in your code to get an image.
The next one is why do you fetch the content from one file and write them to another file to redirect to them? That is not really fast if you write every time the file to another location.
If you get the content echt echo the content and set the correct header to show the image.
header('Content-type:image/png');
readfile($fullpath);
Its much easier and you have a less IO to show files. Otherwise you can use a script like PHPThumb which generated smaller versions and cache the files.
http://phpthumb.sourceforge.net/
I am trying to read multiple image files from a folder (.htaccess protected) and display in a HTML page using php readfile().
The problem is I can see only the first image is read and the next is not shown in the browser. The code is as below
<?php
$image1 = 'files/com_download\256\50\www\res\icon\android\icon-36-ldpi.png';
$image2 = 'files/com_download\256\50\www\res\icon\android\icon-48-mdpi.png';
$imginfo = getimagesize($image1);
header("Content-type: ".$imginfo['mime']);
readfile($image1);
$imginfo = getimagesize($image2);
header("Content-type: ".$imginfo['mime']);
readfile($image2);
?>
I could see the first image 'icon-36-ldpi.png' successfully read and displayed in the browser and the second image is not read and not displayed in the browser.
Am I missing something? Any advice please.
Sorry if I am doing stupid but the requirement is to read multiple image files and render in the browser like a grid view. I cannot use img tag because of security reasons.
You can't dump both images out at once. Why not make two images in your html so the browser makes two calls to your script. Then use a GET param to pass the filename you want to display.
---Edit---
Important Security Note
There is an attack vector which you open up when doing soething like this. Someone could easily view your source html and change the parameter to get your image script to output any file they want. They could even use "../../" to go up directories and search for well known files that exist. e.g. "../../../wp_config.php". Now the attacker has your wordpress database credentials. The correct way to prevent against this is to always validate the input parameter properly. For example, only output if the file name ends with ".jpg"
Following is the piece of code which is not working:-
img src="/old/datagraphsarchive.php?graphing=1&streamCode=ATQ_CALD&date=2008-07-04&iver=1"
Is there any problem with the way I have specified the image source?
This is potentially loading the php file directly instead of as a script. Try using the full external URL in your img src and see if it works.
For example: http://www.yoursite.com/old/datagraphsarchive.php?graphing=1&streamCode=ATQ_CALD&date=2008-07-04&iver=1
If that works, I've yet to actually determine what causes the phenomenon, but you can use the full external url!
If neither work, then your problem lies in the way you're generating your image from the graphsarchive.php file. Ensure you are setting the correct Content-Type headers and writing the binary stream to the page content, like so:
$file = 'myimage.jpg';
header('Content-type: image/jpeg');
readfile($file);
I want to have a PNG picture, but when accessing it, it runs a PHP script, the PHP script should decide what picture to send (using some if statements and whatever). Then the PHP script should read the image file from somewhere on my web server and output it.
Here is the issue, if I get a .png file, and put PHP code in it, it won't work, however, if I use the .php extension, it works, and I can even embed the image into other websites, and the PHP can decide what image to send, but if I want to view that image directly (copy it's URL into my address bar) it doesn't work, it gives me the images plain contents (random jibberish).
Anyone know what to do?
Also This is my first question on Stack Overflow - please tell me if I am doing something wrong.
You need to send Content-Type headers.
For png:
header('Content-Type: image/png');
For others change png to jpg or gif or bmp or whatever.
Please note that header() function must be used before anything is written to output.
First, make sure you have your image image.png somewhere accessible to php.
Then create a php script image.php:
<?php
header('Content-Type: image/png');
readfile('image.png');
The script now acts like it was a PNG image.
It sounds like you know how to send the image, your issue is that you want the URL to look like it's a PNG image.
There are a couple of things you can do. First, if your web server supports URL rewriting (like Apache's mod_rewrite module), you can use a rewrite rule so that the user access the script as something like http://example.com/generated_image.png but your server will translate/rewrite this URL to point directly to your PHP script, so something like /var/www/image_generator.php.
Another option would be to actually name your script "generated_image.png" but force your webserver to treat it like a PHP script. For instance, in Apache you could try something like:
<Location /generated_image.png>
ForceType application/x-httpd-php
</Location>
As a final note, if you're not actually worried about the URL, but worried about the file name that is used if the user decides to save it to disk, you can simply use the Content-Disposition HTTP header in your response. In PHP it would look something like this:
<?php
header("Content-Disposition: inline; filename="generated_image.png");
?>
With that, it doesn't matter what the URL is, if the user saves the image through their web browser, the web browser should offer "generated_image.png" as the default filename.
Simplest version I know...
<?php
header('Content-Type: image/png');
if(whatever)
{
$image=your_image_select_function();
}
// as suggested by sh1ftst0rm with correction of unmatched quotes.
header('Content-Disposition: inline; filename="'.$your_name_variable.'"');
readfile($image);
?>
Then, you treat it like an image file. That is, if this is "pngmaker.php" then, in your HTML document, you do
<img src="pngmaker.php">
You can even do
<img src="pngmaker.php/?id=123&user=me">