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.
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Answer
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.