On (23/12/04 10:40), Cameron Hutchison wrote:
> Once upon a time Clive Menzies said...
> >
> > I've never really understood what the first digit does but
> > having reread the chmod manpage it falls into place almost. Setting
> > the group ID at 2, means any file or directory created by some
Once upon a time Vineet Kumar said...
> * Cameron Hutchison ([EMAIL PROTECTED]) [041222 14:56]:
> > # find $dir -type f -print0 | xargs -0 chmod g=u,o=-rwx
>
> To reduce this yet further, you could do it as
>
> chmod -R g=u $dir
>
> This handles both directories and files in one pass, setting g
* Cameron Hutchison ([EMAIL PROTECTED]) [041222 14:56]:
> # find $dir -type f -print0 | xargs -0 chmod g=u,o=-rwx
To reduce this yet further, you could do it as
chmod -R g=u $dir
This handles both directories and files in one pass, setting group
perms equal to user perms, and nothing else (no m
Hi!
Finally, I could change the default umask. The winner is Cameron Hutchinson
and the libpam-umask package. :-) I downloaded, and compiled it for my
Woody machine (it is only available in Sagre and up), and it works as I
expected.
Many thanks for all your comments!
Alex
Merry Christmas for ever
On Wed, Dec 22, 2004 at 10:54:29PM +, Clive Menzies wrote:
> However, I personally did chmod -R 770 on the basis that there aren't
> usually executables in people's data files (at least not the users I'm
> catering for). Nevetheless I can see the desirability of eliminating
> the possibility o
Once upon a time Clive Menzies said...
>
> I've never really understood what the first digit does but
> having reread the chmod manpage it falls into place almost. Setting
> the group ID at 2, means any file or directory created by someone in 'group'
> will apply the same attributes?
Not q
On (23/12/04 09:51), Cameron Hutchison wrote:
> > On (22/12/04 20:19), Juhasz Sandor wrote:
> > >
> > > I have to implement default umask 002 for my users on my Debian server.
> > > I use KDM login. I searched the net, and I found tips only on setting
> > > umask
> > > on console, and on terminal
Once upon a time Cameron Hutchison said...
>
> To prepare a directory hierarchy for group use, I do the following:
>
> # chgrp -R $group $dir
> # find $dir -type d -print0 | xargs -0 chmod 2770
>
> (I usually use mode 2775, but I think you wanted 2770 from your
> description)
>
> # find $dir -t
On (22/12/04 17:37), David Mandelberg wrote:
> Clive Menzies wrote:
> > For existing directories: $ chmod -R 775 should do the trick
> That will work for the directories themselves, but all files in them will be
> executable.
>
> I wrote a (very basic) bash script that will recursively make normal
> On (22/12/04 20:19), Juhasz Sandor wrote:
> >
> > I have to implement default umask 002 for my users on my Debian server.
> > I use KDM login. I searched the net, and I found tips only on setting umask
> > on console, and on terminal emulators. (The standard /etc/profile,
> > ~/.bashrc, /etc/bas
Menzies
Cc: debian-user@lists.debian.org
Subject: Re: Umask 002 policy
Clive Menzies wrote:
> For existing directories: $ chmod -R 775 should do the trick
That will work for the directories themselves, but all files in them will be
executable.
I wrote a (very basic) bash script that will recursiv
Clive Menzies wrote:
> For existing directories: $ chmod -R 775 should do the trick
That will work for the directories themselves, but all files in them will be
executable.
I wrote a (very basic) bash script that will recursively make normal files one
permission and directories another.[1] Just do
On (22/12/04 20:19), Juhasz Sandor wrote:
> Hi!
>
> I have to implement default umask 002 for my users on my Debian server.
> I use KDM login. I searched the net, and I found tips only on setting umask
> on console, and on terminal emulators. (The standard /etc/profile,
> ~/.bashrc, /etc/bash.bash
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