CODEWITHSUNDEEP
phpimage-processinghttp-headersmime-types
Kladskull
Kladskull

Reputation: 10732

Return a PHP page as an image

I am trying to read a image file (.jpeg to be exact), and 'echo' it back to the page output, but have is display an image...

my index.php has an image link like this:

<img src='test.php?image=1234.jpeg' />

and my php script does basically this:

1) read 1234.jpeg 2) echo file contents... 3) I have a feeling I need to return the output back with a mime-type, but this is where I get lost

Once I figure this out, I will be removing the file name input all together and replace it with an image id.

If I am unclear, or you need more information, please reply.

Upvotes: 79

Views: 123481

Answers (6)

Blaze612 YT
Blaze612 YT

Reputation: 702

Very, very easy.

<?php

//could be image/jpeg or image/gif or whatever
header('Content-Type: image/png')
readfile('image.png')
?>

Upvotes: 5

Boycott A.I.
Boycott A.I.

Reputation: 18871

I feel like we can make this code a little bit easier by just getting the mime type from $image_info:

$file_out = "myDirectory/myImage.gif"; // The image to return

if (file_exists($file_out)) {

   $image_info = getimagesize($file_out);

   //Set the content-type header as appropriate
   header('Content-Type: ' . $image_info['mime']);

   //Set the content-length header
   header('Content-Length: ' . filesize($file_out));

   //Write the image bytes to the client
   readfile($file_out);
}
else { // Image file not found

    header($_SERVER["SERVER_PROTOCOL"] . " 404 Not Found");

}

With this solution any type of image can be processed but it is just another option. Thanks ban-geoengineering for your contribution.

Upvotes: 31

Berk
Berk

Reputation: 75

I worked without Content-Length . maybe reason work for remote image files

// open the file in a binary mode
$name = 'https://www.example.com/image_file.jpg';
$fp = fopen($name, 'rb');

// send the right headers
header('Cache-Control: no-cache, no-store, max-age=0, must-revalidate');
header('Expires: January 01, 2013'); // Date in the past
header('Pragma: no-cache');
header("Content-Type: image/jpg");
/* header("Content-Length: " . filesize($name)); */

// dump the picture and stop the script
fpassthru($fp);
exit;

Upvotes: 5

Drew LeSueur
Drew LeSueur

Reputation: 20145

This should work. It may be slower.

$img = imagecreatefromjpeg($filename);
header("Content-Type: image/jpg");
imagejpeg($img);
imagedestroy($img);

Upvotes: 3

Martin Geisler
Martin Geisler

Reputation: 73748

The PHP Manual has this example:

<?php
// open the file in a binary mode
$name = './img/ok.png';
$fp = fopen($name, 'rb');

// send the right headers
header("Content-Type: image/png");
header("Content-Length: " . filesize($name));

// dump the picture and stop the script
fpassthru($fp);
exit;
?>

The important points is that you must send a Content-Type header. Also, you must be careful not include any extra white space (like newlines) in your file before or after the <?php ... ?> tags.

As suggested in the comments, you can avoid the danger of extra white space at the end of your script by omitting the ?> tag:

<?php
$name = './img/ok.png';
$fp = fopen($name, 'rb');

header("Content-Type: image/png");
header("Content-Length: " . filesize($name));

fpassthru($fp);

You still need to carefully avoid white space at the top of the script. One particularly tricky form of white space is a UTF-8 BOM. To avoid that, make sure to save your script as "ANSI" (Notepad) or "ASCII" or "UTF-8 without signature" (Emacs) or similar.

Upvotes: 135

Joe Bubna
Joe Bubna

Reputation: 296

Another easy Option (not any better, just different) if you aren't reading from a database is to just use a function to output all the code for you... Note: If you also wanted php to read the image dimensions and give that to the client for faster rendering, you could easily do that too with this method.

<?php
  Function insertImage( $fileName ) {
    echo '<img src="path/to/your/images/',$fileName,'">';    
  }
?>

<html>
  <body>
    This is my awesome website.<br>
    <?php insertImage( '1234.jpg' ); ?><br>
    Like my nice picture above?
  </body>
</html>

Upvotes: -7

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