Re: [Bug Report] mod_perl runs with GID 0 !

2016-04-12 Thread John Dunlap
The only stupid question is the one you don't ask. On Tue, Apr 12, 2016 at 10:59 AM, Ruben Safir wrote: > On 04/12/2016 10:54 AM, Ben RUBSON wrote: > > The files created inside a directory will be owned by the group of the > directory, not by the primary group of the user which creates the file.

Re: [Bug Report] mod_perl runs with GID 0 !

2016-04-12 Thread Ruben Safir
On 04/12/2016 10:54 AM, Ben RUBSON wrote: > The files created inside a directory will be owned by the group of the > directory, not by the primary group of the user which creates the file... > This confused me... > > I'm really sorry for the useless noise :| its not noise. I learned something,

Re: [Bug Report] mod_perl runs with GID 0 !

2016-04-12 Thread Ben RUBSON
>> And what is a little bit frightening here is that Apache is launched by the >> #1025 user itself. >> So how could it be possible for it to generate a file owned by group #0... >> Very strange ! > > Something is misconfigured, but its not related to modperl. You're totally right, and I would l

Re: [Bug Report] mod_perl runs with GID 0 !

2016-04-12 Thread John D Groenveld
In message , Ben RUBSON writes: >And what is a little bit frightening here is that Apache is launched by the >#1025 user itself. >So how could it be possible for it to generate a file owned by group #0... >Very strange ! Something is misconfigured, but its not related to modperl. Disable modperl

Re: [Bug Report] mod_perl runs with GID 0 !

2016-04-12 Thread Ben RUBSON
>> Let's run the following example script in mod_perl : >> >> # more test.pl >> open(my $fh, ">", "/tmp/test.log"); >> print $fh "Your UID is " . $< . "\n"; >> my @groups = split '\s', $(; >> print $fh "You belong to these groups: "; >> print $fh $_ . " " foreach(@groups); > > How did you run the

Re: [Bug Report] mod_perl runs with GID 0 !

2016-04-12 Thread Ben RUBSON
>> Let's run the following example script in mod_perl : >> >> # more test.pl >> open(my $fh, ">", "/tmp/test.log"); >> print $fh "Your UID is " . $< . "\n"; >> my @groups = split '\s', $(; >> print $fh "You belong to these groups: "; >> print $fh $_ . " " foreach(@groups); > > How did you run the

Re: [Bug Report] mod_perl runs with GID 0 !

2016-04-12 Thread Ben RUBSON
>> Let's run the following example script in mod_perl : >> >> # more test.pl >> open(my $fh, ">", "/tmp/test.log"); >> print $fh "Your UID is " . $< . "\n"; >> my @groups = split '\s', $(; >> print $fh "You belong to these groups: "; >> print $fh $_ . " " foreach(@groups); > > How did you run the

Re: [Bug Report] mod_perl runs with GID 0 !

2016-04-12 Thread John D Groenveld
In message <88583f53-89d2-40aa-b850-cc819c5e6...@gmail.com>, Ben RUBSON writes: >Let's run the following example script in mod_perl : > ># more test.pl >open(my $fh, ">", "/tmp/test.log"); >print $fh "Your UID is " . $< . "\n"; >my @groups = split '\s', $(; >print $fh "You belong to these groups: "

Re: [Bug Report] mod_perl runs with GID 0 !

2016-04-12 Thread Ben RUBSON
>> Because this is the documented procedure : >> http://perl.apache.org/docs/2.0/user/help/help.html#Reporting_Problems >> > > Interesting. I don't ever recall seeing a bug report on this list. I > always assumed there to be a developed list. I had a look at the Apache BugZilla, but there is n

Re: [Bug Report] mod_perl runs with GID 0 !

2016-04-12 Thread Ruben Safir
On 04/12/2016 02:11 AM, Ben RUBSON wrote: > Because this is the documented procedure : > http://perl.apache.org/docs/2.0/user/help/help.html#Reporting_Problems > Interesting. I don't ever recall seeing a bug report on this list. I always assumed there to be a developed list. > >> I'm confused