bhammel Posted January 23, 2012 Share Posted January 23, 2012 I having been searching through the forum and other websites for information on reading a file (an image file) located above my root directory. I would like to store images outside the root directory and then use PHP to call those images for registered users. Basically I want to create a secure web gallery. I am using GoDaddy Deluxe shared hosting, and have managed to create a webroot directory and place files above it. I have found several scripts for reading files using the readfile() function, however when I actually implement the script to find the file and read its contents, it does not seem to find the file. I know my directory path is correct, as I can call an echo function inside a test.php file located in the same folder as the images (above my root). I would love some incite as to where I went wrong, or if there is a specific GoDaddy work around that needs to be implemented. Also, if I am going about this the wrong way, please let me know if there is a more effective way of going about this! Thank you so much! Quote Link to comment Share on other sites More sharing options...
scootstah Posted January 23, 2012 Share Posted January 23, 2012 Code? Quote Link to comment Share on other sites More sharing options...
bhammel Posted January 23, 2012 Author Share Posted January 23, 2012 Doh! My apologies, I meant to copy the most recent snippet I had tried: <?php $file = basename(urldecode($_GET['file'])) ; $fileDir = '/home/mydomain/photos/'; if (file_exists($fileDir . $file)) { // Note: You should probably do some more checks // on the filetype, size, etc. $contents = file_get_contents($fileDir . $file); // Note: You should probably implement some kind // of check on filetype header('Content-type: image/jpeg'); echo $contents; } ?> This gives me an output of: The image “http://mydomain.com/” cannot be displayed because it contains errors. ---- I substituted mydomain for my real domain, as this is not advertising! Thanks! Quote Link to comment Share on other sites More sharing options...
bhammel Posted January 25, 2012 Author Share Posted January 25, 2012 Well after more searching and trying, I got it to work! This is what I found in case anyone else has this question: $content = file_get_contents('../directory/image.jpg'); /* the directory is actually above the root */ if ($content !== false) { header('Content-type: image/jpeg'); echo $content; } else { echo 'not found'; } This will of course be modified with some SQL search to find the actual image path and then the php file with the above code can be displayed using: <img src="pic.php" /> or so. If anyone has a better suggestion, I would love to optimize this. Thanks! Quote Link to comment Share on other sites More sharing options...
bhammel Posted January 25, 2012 Author Share Posted January 25, 2012 So I have been trying to modify the previous code to work with a dynamically generated file name, and I am having issues with passing variables (it seems). The current code is split between and index.php page that sets a session variable for the image file name, then echos out the file in an image tag using the pic.php file. The pic.php file calls on a function to grab the contents of the file from above my web directory and return those contents to then be output as the pic.php file. If I use the file name (lucas2.jpg) instead of the session variable passed in my function, everything works fine. As soon as I replace the file name with the variable I get broken image. index.php: <?php session_start(); ?> <html> <body> <?php $_SESSION['lucas'] = 'lucas2.jpg'; echo '<img src="pic.php" />'; ?> </body> </html> pic.php: <?php include_once('findimg.php'); $img = findImg($_SESSION['lucas']); if (!empty($img)) { header('Content-type: image/jpeg'); echo $img; } else { echo 'empty img'; } ?> findimg.php: <?php function findImg($image){ $img = $image; $content = file_get_contents('../webpics/'.$img); if ($content !== false) { return $content; } else { echo 'not found'; } } ?> Any thoughts or directions to point me in? Thanks for your time. Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted January 25, 2012 Share Posted January 25, 2012 I don't see a session_start() statement in the pic.php or findimg.php code? Quote Link to comment Share on other sites More sharing options...
bhammel Posted January 25, 2012 Author Share Posted January 25, 2012 I don't see a session_start() statement in the pic.php or findimg.php code? I thought you only need a session_start() function on the first page the session is brought into existence (my index.php page for example). Do I need a session_start() on every page I wish to use session variables? Added session_start() to all pages. Works like a charm. Thank you so much! Quote Link to comment Share on other sites More sharing options...
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