From: Jeffrey Smith <jesmith2095@...>
Date: 2011-09-04T09:35:56+09:00
Subject: Re: opening a file with variable name


On Sep 3, 2011, at 7:39 PM, Hassan Schroeder wrote:

> On Sat, Sep 3, 2011 at 4:26 PM, Jeffrey Smith <jesmith2095@gmail.com> wrote:
>> I changed the slashed and escaped the file path ( I noticed after the fact that my path has spaces), and no change in the error at all.
> 
>>> Again, what OS? <<
> 
> Have you tried 1) opening a file that's in the same directory you've
> started IRB in? 2) changing directories to the one with the target file
> and starting IRB there?
> 
> -- 
> Hassan Schroeder ------------------------ hassan.schroeder@gmail.com
> http://about.me/hassanschroeder
> twitter: @hassan
> 

The OS is Windows 2003.

I can open the file from the cmd line (not IRB), and when I hardcode the path in File.open(\\\\sdcfaxgw04\\c$\\epic\\jobs\\Processed\\7.7.7\\Epic Print Service\\xxx) it works fine. 

I created the variable by combining two other variables, and when I try the File.exists? it fails. If I assign the complete path including file (i.e. \\\\sdcfaxgw04\\c$\\epic\\jobs\\Processed\\7.7.7\\Epic Print Service\\sd3n4v1.20110902.1924.55435838.JMym.epic) it works fine.

Could it be the way I'm creating the variable? This is the process: 

ertf = "\\\\sdcfaxgw04\\c$\\epic\\jobs\\Processed\\7.7.7\\Epic Print Service\\"+filevariable

Jeff