Output an Image in PHP

Question

I have an image $file (eg. ../image.jpg)

which has a mime type $type

How can I output it to the browser?

Answer 1

<?php

header("Content-Type: $type");
readfile($file);

That's the short version. There's a few extra little things you can do to make things nicer, but that'll work for you.

Answer 2

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

Answer 3

header('Content-type: image/jpeg');
readfile($image);

Answer 4

You can use header to send the right Content-type :

header('Content-Type: ' . $type);

And readfile to output the content of the image :

readfile($file);

And maybe (probably not necessary, but, just in case) you'll have to send the Content-Length header too :

header('Content-Length: ' . filesize($file));

Note : make sure you don't output anything else than your image data (no white space, for instance), or it will no longer be a valid image.

Answer 5

Try this:

<?php
  header("Content-type: image/jpeg");
  readfile("/path/to/image.jpg");
  exit(0);
?>

Answer 6

If you have the liberty to configure your webserver yourself, tools like mod_xsendfile (for Apache) are considerably better than reading and printing the file in PHP. Your PHP code would look like this:

header("Content-type: $type");
header("X-Sendfile: $file"); # make sure $file is the full path, not relative
exit();

mod_xsendfile picks up the X-Sendfile header and sends the file to the browser itself. This can make a real difference in performance, especially for big files. Most of the proposed solutions read the whole file into memory and then print it out. That's OK for a 20kbyte image file, but if you have a 200 MByte TIFF file, you're bound to get problems.