Return a PHP Page as an Image

Return a PHP page as an image

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.

How to return image from php file set as img src

filesize does not work on HTTP URLs. The docs say:

This function can also be used with some URL wrappers. Refer to Supported Protocols and Wrappers to determine which wrappers support stat() family of functionality.

However, the HTTP wrapper does not support the stat function. Because of this, you send a wrong Content-Length header and the HTTP response cannot be interpreted by your browser.

I see two possible solutions:

1. Load the image into memory and use strlen:

   $image = file_get_contents('http://example.com/img.jpg');
   
   header('Content-type: image/jpeg;');
   header("Content-Length: " . strlen($image));
   echo $image;
   

2. Use the $http_response_header variable to read the remote response's Content-Length header:

   $img = 'http://example.com/img.jpg';
   $fp = fopen($img, 'rb');
   
   header('Content-type: image/jpeg;');
   foreach ($http_response_header as $h) {
       if (strpos($h, 'Content-Length:') === 0) {
           header($h);
           break;
       }
   }
   
   fpassthru($fp);
   

How to return image

header("Content-type: image/jpeg");
imagejpeg($img);

Return the contents of an image in php file?

Check out readfile().

The basic idea is you send the appropriate MIME type headers (using header()) then deliver the file contents using readfile().

For example

<?php
// myhumbleimage.php

// Do whatever myhumbleimage.php does before the image is delivered

header('Content-Type: image/jpeg');
readfile('path/or/url/of/image/file.jpg');
exit;

Is there any php function that returns an image handle so that i can use it in img src html tag?

There are 2 more obvious ways to approach this problem.

1. Write the rendered image to a file

   This is possible using imagejpeg, by passing in a second parameter of the filename, for example:

    $im = imagecreatetruecolor(..);
    // ..
    imagejpeg($im, 'image.jpg');
   

2. Use URL Rewriting

   You can always pass through all .jpgs in a special folder (for example /dynamic/image.jpg through to your script which will render it). If you want, you could even take the filename (image.jpg) as a parameter to your script telling it what to render if the image is dynamic.

Obviously, if your image needs to be different per request, #2 is the better option. However, #1 is faster and recommended if your image is static (ie. your imagettftext always writes the same text over the same image).

Output an Image in PHP

$file = '../image.jpg';
$type = 'image/jpeg';
header('Content-Type:'.$type);
header('Content-Length: ' . filesize($file));
readfile($file);

How to return an image created with php

imagepng($image), with no second parameter, will simply dump the raw PNG data out to the client (e.g. the browser). No need to write it to a file, then read that file back in.

This is clearly documented in the man page: http://php.net/imagepng and applies to ALL of the "save" functions in GD.

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