Issue with uploading files

Question

leviathan185 14 Junior Poster

14 Years Ago

Hi,

I am trying to write an upload script and everytime i try and upload a file it will give me an error.

Warning: move_uploaded_file(/home/******/public_html/dw/uploads) [function.move-uploaded-file]: failed to open stream: Is a directory in /home/******/public_html/dw/upload2.php on line 3

I have checked that the folder is there as the error suggests that it is not there but it is definately there.

My code is as follows

upload1.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
	<form enctype="multipart/form-data" method="post" action="upload2.php">
    	Send this file: <input name="userfile" type="file" /><br  />
        <input type="submit" value="Send File" />
    </form>
</body>
</html>

and upload2.php

<?php
					
	if (move_uploaded_file($_FILES['userfile']['tmp_name'], "/home/alanhuno/public_html/dw/uploads")) {
		print "Received {$_FILES['userfile']['name']} - its size is {$_FILES['userfile']['size']}";
	} else {
		print "Upload Failed";
	}
?>

php

6 Contributors
16 Replies
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Latest Post 14 Years Ago Latest Post by leviathan185

All 16 Replies

jcacquiescent27 0 Light Poster

14 Years Ago

I've experienced an error message similar to this when dealing with the file system. At the time I had the problem, the solution came in the form of checking the folder permissions to be sure the proper write access existed.

praghna 0 Newbie Poster

14 Years Ago

check weather the path given is correct or not.folder permissions are set automatically.

cwarn23 387 Occupation: Genius

14 Years Ago

Try making a path relative to your script like the following:

<?php
					
	if (move_uploaded_file($_FILES['userfile']['tmp_name'], "uploads/")) {
		print "Received {$_FILES['userfile']['name']} - its size is {$_FILES['userfile']['size']}";
	} else {
		print "Upload Failed";
	}
?>

Also you can try this altered version:

<?php
					
	if (move_uploaded_file($_FILES['userfile']['tmp_name'], "/home/alanhuno/public_html/dw/uploads/")) {
		print "Received {$_FILES['userfile']['name']} - its size is {$_FILES['userfile']['size']}";
	} else {
		print "Upload Failed";
	}
?>

Just note that there needs to be a slash at the end of the path.

cwarn23 387 Occupation: Genius

14 Years Ago

if i use "/uploads/" it says permission denied

That refers to the root when it begins with a slash. So try removing the slash at the beginning so it looks like "uploads/" just like in my sample code.

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leviathan185 14 Junior Poster · Answer 1 · 2009-07-18T06:02:12+00:00

I've experienced an error message similar to this when dealing with the file system. At the time I had the problem, the solution came in the form of checking the folder permissions to be sure the proper write access existed.

Yeah I did check that as well. The folder I am trying to upload to has 0777 (Full Access).

leviathan185 14 Junior Poster · Answer 2 · 2009-07-18T12:36:21+00:00

check weather the path given is correct or not.folder permissions are set automatically.

I have also tried using the following for makng sure the path is correct.

$thisDir = getcwd();
$sveToDir = $thisDir ."/uploads";

	if (move_uploaded_file($_FILES['userfile']['tmp_name'], "$saveToDir")) {		print "Received {$_FILES['userfile']['name']} - its size is {$_FILES['userfile']['size']}";	} else {		print "Upload Failed";	}

leviathan185 14 Junior Poster · Answer 3 · 2009-07-20T03:35:09+00:00

Try making a path relative to your script like the following:

<?php
					
	if (move_uploaded_file($_FILES['userfile']['tmp_name'], "uploads/")) {
		print "Received {$_FILES['userfile']['name']} - its size is {$_FILES['userfile']['size']}";
	} else {
		print "Upload Failed";
	}
?>

Also you can try this altered version:

<?php
					
	if (move_uploaded_file($_FILES['userfile']['tmp_name'], "/home/alanhuno/public_html/dw/uploads/")) {
		print "Received {$_FILES['userfile']['name']} - its size is {$_FILES['userfile']['size']}";
	} else {
		print "Upload Failed";
	}
?>

Just note that there needs to be a slash at the end of the path.

even with the / at the end this is still giving me the same thing. i have contacted my server support and they advsied that the permissions need to be set to 775 as they do not allow 777 access. so i now know that the permissions are correct.

at this point if i use the full path it says its not there
if i use "/uploads/" it says permission denied

I swear i am going to go bald at 24... AHGHHHH

leviathan185 14 Junior Poster · Answer 4 · 2009-07-21T01:42:05+00:00

That refers to the root when it begins with a slash. So try removing the slash at the beginning so it looks like "uploads/" just like in my sample code.

even with out the / at the start is is giving:

Warning: move_uploaded_file(uploads/) [function.move-uploaded-file]: failed to open stream: Is a directory in /home/alanhuno/public_html/upload2.php on line 2

Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move '/var/tmp/phpIfPXPI' to 'uploads/' in /home/alanhuno/public_html/upload2.php on line 2
Upload Failed

i even moved all the files to the root folder to see if that would work.

leviathan185 14 Junior Poster · Answer 5 · 2009-07-21T01:48:11+00:00

the form to upload the file is

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form enctype="multipart/form-data" method="post" action="upload2.php">
    Send this file: <input name="userfile" type="file" /><br />
    <input type="submit" value="Send File" />
</form> 
</body>
</html>

I am pretty sure that is correct too but thought id post just in case.

mschroeder 251 Bestower of Knowledge Team Colleague · Answer 6 · 2009-07-21T02:37:24+00:00

The problem you're having is $_FILES is an actual filename. your destination is a directory without a file name. its trying to create a file named "uploads", which is a valid filename on a linux system, in your dw folder. Try actually giving the file a filename in the destination path.

Example:

<?php
$uploads_dir = '/uploads';
foreach ($_FILES["pictures"]["error"] as $key => $error) {
    if ($error == UPLOAD_ERR_OK) {
        $tmp_name = $_FILES["pictures"]["tmp_name"][$key];
        $name = $_FILES["pictures"]["name"][$key];
        move_uploaded_file($tmp_name, "$uploads_dir/$name");
    }
}
?>

Source: php.net (move_uploaded_file)

Notice how in that code the destination is /uploads/[name of the actual file]. Should solve your problem.

leviathan185 14 Junior Poster · Answer 7 · 2009-07-21T12:36:02+00:00

The problem you're having is $_FILES is an actual filename. your destination is a directory without a file name. its trying to create a file named "uploads", which is a valid filename on a linux system, in your dw folder. Try actually giving the file a filename in the destination path.
Example:
<?php
$uploads_dir = '/uploads';
foreach ($_FILES["pictures"]["error"] as $key => $error) {
    if ($error == UPLOAD_ERR_OK) {
        $tmp_name = $_FILES["pictures"]["tmp_name"][$key];
        $name = $_FILES["pictures"]["name"][$key];
        move_uploaded_file($tmp_name, "$uploads_dir/$name");
    }
}
?>
Source: php.net (move_uploaded_file)
Notice how in that code the destination is /uploads/[name of the actual file]. Should solve your problem.

I tried the above code but it gives an error on the foreach line saying that it is an invalid argument?

i tried modifiying the code to suit my form code and still got the same error

<?php
	$uploads_dir = "uploads/";
	
	foreach($_FILES['userfile']['error'] as $key => $error) {
	
		if($error == UPLOAD_ERR_OK) {
			
			$tmp_name = $_FILES['userfile']['tmp_name'][$key];
			$name = $_FILES['userfile']['name'][$key];
			move_uploaded_file($tmp_name, $uploads_dir .$name);
		}
	}	
?>

The error i got is:
Warning: Invalid argument supplied for foreach() in /home/alanhuno/public_html/upload2.php on line 4

leviathan185 14 Junior Poster · Answer 8 · 2009-07-21T12:56:16+00:00

I tried the above code but it gives an error on the foreach line saying that it is an invalid argument?
i tried modifiying the code to suit my form code and still got the same error
<?php
	$uploads_dir = "uploads/";
	
	foreach($_FILES['userfile']['error'] as $key => $error) {
	
		if($error == UPLOAD_ERR_OK) {
			
			$tmp_name = $_FILES['userfile']['tmp_name'][$key];
			$name = $_FILES['userfile']['name'][$key];
			move_uploaded_file($tmp_name, $uploads_dir .$name);
		}
	}	
?>
The error i got is:
Warning: Invalid argument supplied for foreach() in /home/alanhuno/public_html/upload2.php on line 4

I managed to get it working when i changed the code to

<?php
	$uploads_dir = "uploads/";
	
	$tmp_name = $_FILES['userfile']['tmp_name'];
	$name = $_FILES['userfile']['name'];
	move_uploaded_file($tmp_name, $uploads_dir .$name)
		or die("could not upload file");
?>

Thanks to all who tried to help. I hope my hair grows back now :P

basantxbs 0 Newbie Poster · Answer 9 · 2009-07-21T16:10:56+00:00

Hi,
I am trying to write an upload script and everytime i try and upload a file it will give me an error.
Warning: move_uploaded_file(/home/******/public_html/dw/uploads) [function.move-uploaded-file]: failed to open stream: Is a directory in /home/******/public_html/dw/upload2.php on line 3
I have checked that the folder is there as the error suggests that it is not there but it is definately there.
My code is as follows
upload1.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
	<form enctype="multipart/form-data" method="post" action="upload2.php">
    	Send this file: <input name="userfile" type="file" /><br  />
        <input type="submit" value="Send File" />
    </form>
</body>
</html>
and upload2.php
<?php
					
	if (move_uploaded_file($_FILES['userfile']['tmp_name'], "/home/alanhuno/public_html/dw/uploads")) {
		print "Received {$_FILES['userfile']['name']} - its size is {$_FILES['userfile']['size']}";
	} else {
		print "Upload Failed";
	}
?>

are u working with xamp or wamp ?
this is only the path problem rest of the code is ok.

mschroeder 251 Bestower of Knowledge Team Colleague · Answer 10 · 2009-07-21T18:30:50+00:00

That is what I was trying to illustrate as well, that it was a path problem, aka the filename was missing from the path, the code itself was fine. The example i posted was from the manual illustrating path + filename. I apologize for any confusion caused to the op.

basantxbs 0 Newbie Poster · Answer 11 · 2009-07-22T15:51:58+00:00

$img_name=time().$_FILES['img']['name']; //  allow to upload more than one file with same name     
$temp_file=$_FILES['img']['tmp_name'];
		$dest_path="Image/".$img_name;
		
                if(move_uploaded_file($temp_file,$dest_path)){
                    echo "File is sucessfully uploaded at Image directory";
                }else{
                    echo "Error in uploading file";
                }

You can use this code . This will store your file at image directory.

leviathan185 14 Junior Poster · Answer 12 · 2009-07-23T02:47:06+00:00

$img_name=time().$_FILES['img']['name']; //  allow to upload more than one file with same name     
$temp_file=$_FILES['img']['tmp_name'];
		$dest_path="Image/".$img_name;
		
                if(move_uploaded_file($temp_file,$dest_path)){
                    echo "File is sucessfully uploaded at Image directory";
                }else{
                    echo "Error in uploading file";
                }

Thanks that is a great idea, I was just going to check if the file exists and add a version number to it but that sounds a lot easier. thanks

Issue with uploading files

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