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PHP Upload Script

Discussion in 'PHP' started by dean5000v, May 1, 2008.

0
  1. #1
    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:

     
    dean5000v, May 1, 2008 IP
  2. CPURules

    CPURules Peon

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    #2 
    
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
     
    CPURules, May 1, 2008 IP
  3. dean5000v

    dean5000v Peon

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    #3
    nope wasn't confusing at all, thanks a lot for the example !!
     
    dean5000v, May 1, 2008 IP
  4. dean5000v

    dean5000v Peon

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    #4
    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
     
    dean5000v, May 1, 2008 IP
  5. dean5000v

    dean5000v Peon

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    #5
    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:
     
    dean5000v, May 1, 2008 IP
  6. DarkMindZ

    DarkMindZ Guest

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    #6
    DarkMindZ, May 1, 2008 IP