Question

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.

Answer 1

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.