copy() function help

[wchamber22]

Hello freaks,

 

I am currently creating a member system that will place a default handle image into a newly registered members file.

I have used the mkdir function create the directory successfully based on the members ID, but I can't get the copy function to work.

 

Code:

mkdir("memberFiles/$memberID", 0755); 
$old = 'images/defaultlogo.jpg';
$new = 'memberFiles/$memberID/logo.jpg';
copy($old, $new);

 

Error Message:

Warning: copy(memberFiles/$memberID/logo.jpg) [function.copy]: failed to open stream: No such file or directory

 

Lastly, this folder is created to store members files.

[AyKay47]

this doesn't have to do with $new, as it will be overwritten even if it did exist. Have you read the error? It's telling you that it cannot locate the file specified in $old. Perhaps check for the proper path.

[kicken]

Variables are not translated when used in single-quoted strings.  So your trying to copy your image to a directory literally named '$memberID' rather than one with that ID value.

 

Use double quotes or concatenate.

 

[DavidAM]

Read the error message carefully, and you will find that the variable name is in it ($memberID).

Warning: copy(memberFiles/$memberID/logo.jpg) [function.copy]: failed to open stream: No such file or directory

 

Variables are not interpreted inside single-quote marks. Use double-quotes to assign the the "old" name:

$new = "memberFiles/$memberID/logo.jpg";

The error message could be a little clearer; since it shows only one of the two parameters, it lead me to believe that you had them backwards. But, I suspect the error indicates that the directory "$memberID" does not exist and therefore you cannot copy a file into it.

[AyKay47]

ah, good catch.

[wchamber22]

Thank you ALL,

 

When I switched the singles to doubles and all was well.  I am pretty new to this, but am infinately better than when I started, thanks to these forums!