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

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

Answer 2

For the next guy or gal hitting this problem, here's what worked for me:

ob_start();
header('Content-Type: '.$mimetype);
ob_end_clean();
$fp = fopen($fullyQualifiedFilepath, 'rb');
fpassthru($fp);
exit;

You need all of that, and only that. If your mimetype varies, have a look at PHP's mime_content_type($filepath)

Answer 3

<?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 4

$file = '../image.jpg';

if (file_exists($file))
{
    $size = getimagesize($file);

    $fp = fopen($file, 'rb');

    if ($size and $fp)
    {
        // Optional never cache
    //  header('Cache-Control: no-cache, no-store, max-age=0, must-revalidate');
    //  header('Expires: Mon, 26 Jul 1997 05:00:00 GMT'); // Date in the past
    //  header('Pragma: no-cache');

        // Optional cache if not changed
    //  header('Last-Modified: '.gmdate('D, d M Y H:i:s', filemtime($file)).' GMT');

        // Optional send not modified
    //  if (isset($_SERVER['HTTP_IF_MODIFIED_SINCE']) and 
    //      filemtime($file) == strtotime($_SERVER['HTTP_IF_MODIFIED_SINCE']))
    //  {
    //      header('HTTP/1.1 304 Not Modified');
    //  }

        header('Content-Type: '.$size['mime']);
        header('Content-Length: '.filesize($file));

        fpassthru($fp);

        exit;
    }
}

http://php.net/manual/en/function.fpassthru.php

Answer 5

^{(Expanding on the accepted answer...)}

I needed to:

1. log views of a jpg image and an animated gif, and,
2. ensure that the images are never cached (so every view is logged), and,
3. also retain the original file extensions.

---

I accomplished this by creating a "secondary" .htaccess file in the sub-folder where the images are located.
The file contains only one line:

AddHandler application/x-httpd-lsphp .jpg .jpeg .gif

In the same folder, I placed the two 'original' image files (we'll call them orig.jpg and orig.gif), as well as two variations of the [simplified] script below (saved as myimage.jpg and myimage.gif)...

<?php 
  error_reporting(0); //hide errors (displaying one would break the image)

  //get user IP and the pseudo-image's URL
  if(isset($_SERVER['REMOTE_ADDR'])) {$ip =$_SERVER['REMOTE_ADDR'];}else{$ip= '(unknown)';}
  if(isset($_SERVER['REQUEST_URI'])) {$url=$_SERVER['REQUEST_URI'];}else{$url='(unknown)';}

  //log the visit
  require_once('connect.php');            //file with db connection info
  $conn = new mysqli($servername, $username, $password, $dbname);
  if (!$conn->connect_error) {         //if connected then save mySQL record
   $conn->query("INSERT INTO imageclicks (image, ip) VALUES ('$url', '$ip');");
     $conn->close();  //(datetime is auto-added to table with default of 'now')
  } 

  //display the image
  $imgfile='orig.jpg';                             // or 'orig.gif'
  header('Content-Type: image/jpeg');              // or 'image/gif'
  header('Content-Length: '.filesize($imgfile));
  header('Cache-Control: no-cache');
  readfile($imgfile);
?>

The images render (or animate) normally and can be called in any of the normal ways for images (like an <img> tag), and will save a record of the visiting IP, while invisible to the user.

Answer 6

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

This is works for me! I have test it on code igniter. if i use readfile, the image won't display. Sometimes only display jpg, sometimes only big file. But after i changed it to "file_get_contents" , I get the flavour, and works!! this is the screenshoot: Screenshot of "secure image" from database