From: Steve Grubb <sgrubb@redhat.com>
To: linux-audit@redhat.com
Subject: Re: linux auditd: Not getting log for chmod syscall
Date: Fri, 13 Jan 2012 15:04:37 -0500 [thread overview]
Message-ID: <201201131504.38174.sgrubb@redhat.com> (raw)
In-Reply-To: <CAKYigEA5LK_RdgY8-ae-H7yMh_mZKNg-R7b4SnfMLKTJttnoRA@mail.gmail.com>
On Thursday, January 12, 2012 11:52:29 PM bharat gupta wrote:
> I am using redhat 6, and trying to create logs for some system call using
> the rule given below:
>
> *-a always,exit -F arch=b64 -S chmod -S fchmod -S fchmodat -F auid>=500
> -F auid!=4294967295 -k perm_mod*
The rule works for me.
# auditctl -a always,exit -F arch=b64 -S chmod -S fchmod -S fchmodat -F
'auid>=500' -F auid!=4294967295 -k perm_mod
I don't have any asterisk and I have single quote marks since bash will
interpret the > as a redirection. But then doing a chmod command, it does pick
up the fchmodat() syscall.
> After running command chmod i was not able to get any log, but when i used
> strace command i have seen that syscall have been called.
> I also checked that auditd service is running properly.
When you use auditctl -l, is the rule just like you expected?
LIST_RULES: exit,always arch=3221225534 (0xc000003e) auid>=500 (0x1f4) auid!=-1
(0xffffffff) key=perm_mod syscall=chmod,fchmod,fchmodat
It should just work unless you are on a distribution that does not really
support auditing.
-Steve
next prev parent reply other threads:[~2012-01-13 20:04 UTC|newest]
Thread overview: 6+ messages / expand[flat|nested] mbox.gz Atom feed top
2012-01-13 4:52 linux auditd: Not getting log for chmod syscall bharat gupta
2012-01-13 20:04 ` Steve Grubb [this message]
2012-01-18 11:10 ` bharat gupta
2012-01-18 12:10 ` Marcelo Cerri
[not found] ` <CAKYigEAYpkm99o1XbhEAz0CrsMFSLBQdp8cY0TCAZxpVzZ1DMw@mail.gmail.com>
[not found] ` <4F17FB57.9010804@linux.vnet.ibm.com>
[not found] ` <CAKYigEDBechU7a=fdf0_aPuK01k2yESx5J7SWcAt2X6qn2pzvA@mail.gmail.com>
[not found] ` <4F180497.2080900@linux.vnet.ibm.com>
[not found] ` <CAKYigEDrCdWVhT7wX4260xe2sUtkm0dd0DuhPfuhHvq98on41Q@mail.gmail.com>
[not found] ` <4F1808F7.1010709@linux.vnet.ibm.com>
[not found] ` <CAKYigEA1eti=0xsgKiyzOavHg6DnjF4pVLGbCj4HvQZ4ViieOw@mail.gmail.com>
[not found] ` <CAKYigEC8zaqkOAOZK6YNzdLqK+9fXbFrVS_0jA=CVsdM9qyMmg@mail.gmail.com>
[not found] ` <4F1EA802.1090003@linux.vnet.ibm.com>
[not found] ` <CAKYigEC-Av7f+0n2zTADiEdNdWzt3QcOC13SnsUn2QodUytWng@mail.gmail.com>
[not found] ` <4F1ECDD2.5040907@linux.vnet.ibm.com>
2012-01-24 15:30 ` Fwd: " bharat gupta
2012-01-24 16:03 ` Steve Grubb
Reply instructions:
You may reply publicly to this message via plain-text email
using any one of the following methods:
* Save the following mbox file, import it into your mail client,
and reply-to-all from there: mbox
Avoid top-posting and favor interleaved quoting:
https://en.wikipedia.org/wiki/Posting_style#Interleaved_style
* Reply using the --to, --cc, and --in-reply-to
switches of git-send-email(1):
git send-email \
--in-reply-to=201201131504.38174.sgrubb@redhat.com \
--to=sgrubb@redhat.com \
--cc=linux-audit@redhat.com \
/path/to/YOUR_REPLY
https://kernel.org/pub/software/scm/git/docs/git-send-email.html
* If your mail client supports setting the In-Reply-To header
via mailto: links, try the mailto: link
Be sure your reply has a Subject: header at the top and a blank line
before the message body.
This is a public inbox, see mirroring instructions
for how to clone and mirror all data and code used for this inbox