Code Help For Uploading Image files

jmkeane

Hi,

I was given the php scripts for this website and there is an error in the one that allows a user to upload an image file from their computer to a table called "photo" in my website. Below is the code:

?
include("cookie.php");
include("connect.php");

if($Error1==1)
$Error="Size of photo can not be more than 400kb";
if($submit)
{
if($photo=="" || $quote=="")
{
$Error="Incomplete Information";
}
else
{
if($photo_size<=409600)
{
$resmaxid=mysql_query("select max(id) from photo");
$rowmaxid=mysql_fetch_row($resmaxid);
$maxid=$rowmaxid[0];

		$PRODUCTIMAGES="photoimage";
		$path =$PRODUCTIMAGES."/".$maxid.$photo_name;
		$photoname=$maxid.$photo_name;
		copy($photo,$path);
		if($fromwhere==1)
		{
			$photoquery="insert into photo(photo,quote,username,main) values('$photoname','$quote','$UsErCoOkIe','Y')";
		}
		else
		{
			$photoquery="insert into photo(photo,quote,username,main) values('$photoname','$quote','$UsErCoOkIe','$maxid')";
		}
		$photoresult=mysql_query($photoquery,$db);
	}
	else
	{
		$Error="Size of photo can not be more than 400kb";
	}
}

}
if($photoresult)
{
header("Location:message.php?11");
}
$sql="select * from users where identity='$UsErCoOkIe'";
$result=mysql_query($sql,$db);
$row7=mysql_fetch_array($result);
$id=$row7["userid"];
?>

the error says "warning: unable to open 'array' for reading no such file or directory.....on line 24"

line 24 is

copy($photo,$path);

What does this error mean? and any suggestions as to how to fix it please?

Cheers,
Joel

Weedpacket

First of all, this must be really old code because it relies on register_globals being turned on (it shouldn't be). Best practice has been this for some years now.

Second, it would appear that $photo is an array; it's been so long since I've used that method of handling file uploads I forget what it is you do want; my recommendation would be to update the code to current standards.

$UsErCoOkIe

Someone is yanking your chain, right?

leatherback

Hi Joel,

Brave to work on someone elses code ;-)

Here: copy($photo,$path);

You copy the uploaded file to a new location. I do not know where $photo is defined, as you do not give us this information. (As a matter of fact.. You seem to have left out a substantial piece of coding, as you do not show at all how you get a whole bunch of variables. If no code is missing, you might want to look at trouble you can run into when you are working with 'register_globals' switched ON.

The problem is, that the $photo variable is not a plain string, but an array. As I am not sure how this gets defined, I cannot tell you exactly how to fix it. However, from the $FILES array (Which is the result of a file-upload box) you should have the uploaded filepath in $FILES['userfile']['tmp_name']

see: http://uk.php.net/manual/en/features.file-upload.php