ok ive done this php upload script, which simply uploads the file to my uploads directory, i wont to be able to add into a function where it sets the a link to the image into a variable and then prints it. here is my code <?php $maxsize=28480; //set the max upload size in bytes if (!$HTTP_POST_VARS['submit']) { //print_r($HTTP_POST_FILES); $error=" "; //this will cause the rest of the processing to be skipped //and the upload form displays } if (!is_uploaded_file($HTTP_POST_FILES['upload_file']['tmp_name']) AND !isset($error)) { $error = "<b>You must upload a file!</b><br /><br />"; unlink($HTTP_POST_FILES['upload_file']['tmp_name']); } if ($HTTP_POST_FILES['upload_file']['size'] > $maxsize AND !isset($error)) { $error = "<b>Error, file must be less than $maxsize bytes.</b><br /><br />"; unlink($HTTP_POST_FILES['upload_file']['tmp_name']); } if($HTTP_POST_FILES['upload_file']['type'] != "image/gif" AND $HTTP_POST_FILES['upload_file']['type'] != "image/pjpeg" AND $HTTP_POST_FILES['upload_file']['type'] !="image/jpeg" AND !isset($error)) { $error = "<b>You may only upload .gif or .jpeg files.</b><br /><br />"; unlink($HTTP_POST_FILES['upload_file']['tmp_name']); } if (!isset($error)) { move_uploaded_file($HTTP_POST_FILES['upload_file']['tmp_name'], "uploads/".$HTTP_POST_FILES['upload_file']['name']); print "thank you for your upload."; exit; } else { echo ("$error"); } ?> <html> <head></head> <body> <form action="<?php echo(htmlspecialchars($_SERVER['PHP_SELF']))?>" method="post" enctype="multipart/form-data"> Choose a file to upload:<br /> <input type="file" name="upload_file" size="80"> <br /> <input type="submit" name="submit" value="submit"> </form> </body> </html> PHP:
if (!isset($error)) { move_uploaded_file($HTTP_POST_FILES['upload_file']['tmp_name'], "uploads/".$HTTP_POST_FILES['upload_file']['name']); print "thank you for your upload."; exit; } PHP: That's part of your script. It needs to be this: if (!isset($error)) { move_uploaded_file($HTTP_POST_FILES['upload_file']['tmp_name'], "uploads/".$HTTP_POST_FILES['upload_file']['name']); print "thank you for your upload.<br />"; print "<a href=\"uploads/" . $HTTP_POST_FILES['upload_file']['tmp_name'] . "\">View your image</a>"; print "(or access it directly: http://www.YOURWEBSITENAME.com/uploads/" . $HTTP_POST_FILES['upload_file']['tmp_name'] . ")"; exit; } PHP: Obviously, change YOURWEBSITENAME.COM to your website, and if this upload script is not in the root DIR of the website, you'll need to change YOURWEBSITENAME.COM to YOURWEBSITENAME.COM/DIR, where DIR is the directory that this script resides in. Sorry if that was a bit confusing :3
ive tried adding in the script and the first link is just basically a link to the uploads file which i don't want, and the second one just gives me a link of the temp file php has given it which doesn't work. the script uploads the file to my uploads file but i just want it so gives me a link the the exact file name in the folder e.g. uploads/image1.jpg mybe im just being stupid though
sorry for doing a third post but i have figured it, wasnt hard im just crap lol. heres the code. print "<a href=\"http://www.mydomain.com/uploads/" . $HTTP_POST_FILES['upload_file']['name'] . "\">View your image</a>"; PHP: