[OpenAFS] afs directory

[Michael Norwick]

Rodney M Dyer wrote:

> At 04:45 PM 7/9/2005, you wrote:
>
>> Is there a method for finding, or list of, the contents of public afs 
>> cells?
>
>
> I'm not sure what you mean.  If you mean simply to perform a "dir /s 
> /b" (windows speak), or "ls -al" (unix speak) of any AFS cell, then 
> yes, it is quite simple.  You just go to "www.openafs.org" and install 
> the OpenAFS client for your operating system.  All AFS cells that 
> aren't behind firewalls are public, at least if the organizations 
> running the cells have published their cell server addresses in the 
> CellServDB (included with the OpenAFS client).  Once you have your 
> OpenAFS client working and an AFS cell is mounted, then you should be 
> able to use "dir" or "ls" to look inside any cell.  However you won't 
> be able to "see" anything you aren't authorized to see.  You will only 
> be able to list and open the contents of directories that have 
> "system:anyuser rl"  (rl -short for read, look).
>
>> Where can I find the current krb-afs source code within the afs network?
>
>
> AFS and Kerberos 5 are separate code bases, and neither is "in" the 
> AFS network.  You can find the source code for AFS at 
> "www.openafs.org".  You can find MIT's Kerberos client and server 
> source code at  http://web.mit.edu/network/kerberos-form.html
>
> Hope this helped.
>
> Rodney
>
>
>
Well, not quite.  I've had a test cell up and compiled several versions 
of OpenAFS, my network uses kerberos authentication to ThisCell via 
krb524.  I just
put together another AFS client using 1.3.84 and while testing I was 
noodling around the various cells and was wondering whether there was 
any easier
way to find the contents of public cells other than ls(ing) their 
contents.  Being a purist I wanted to find the migration kit via AFS 
rather than googling
around the web.  Considered a Perl script but I am not sure how many 
sysadmins I would piss off searching their archives this way (I try to 
be a good
netizen).  So, there it is.
Thanks,
Michael