Bug in Mailman version 2.1.9
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We receive this error: Bug in Mailman version 2.1.9 We're sorry, we hit a bug! Please inform the webmaster for this site of this problem. Printing of traceback and other system information has been explicitly inhibited, but the webmaster can find this information in the Mailman error logs. When attempting to login to the admin interface and the logs say: May 24 17:32:05 2012 admin(20758): @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ admin(20758): [----- Mailman Version: 2.1.9 -----] admin(20758): [----- Traceback ------] admin(20758): Traceback (most recent call last): admin(20758): File "/usr/lib/mailman/scripts/driver", line 101, in run_main admin(20758): main() admin(20758): File "/usr/lib/mailman/Mailman/Cgi/admindb.py", line 251, in main admin(20758): mlist.Save() admin(20758): File "/usr/lib/mailman/Mailman/MailList.py", line 565, in Save admin(20758): self.CheckHTMLArchiveDir() admin(20758): File "/usr/lib/mailman/Mailman/Archiver/Archiver.py", line 241, in CheckHTMLArchiveDir admin(20758): makelink(privdir, pubdir) admin(20758): File "/usr/lib/mailman/Mailman/Archiver/Archiver.py", line 49, in makelink admin(20758): os.symlink(old, new) admin(20758): OSError: [Errno 13] Permission denied admin(20758): [----- Python Information -----] admin(20758): sys.version = 2.4.3 (#1, Sep 3 2009, 15:37:37) [GCC 4.1.2 20080704 (Red Hat 4.1.2-46)] admin(20758): sys.executable = /usr/bin/python admin(20758): sys.prefix = /usr admin(20758): sys.exec_prefix = /usr admin(20758): sys.path = /usr admin(20758): sys.platform = linux2 admin(20758): [----- Environment Variables -----] and the permissions are ls -ld /var/lib/mailman/archives/private drwxrws--x 471 mailman mailman 36864 May 24 17:27 /var/lib/mailman/archives /private Help! thanks
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On 5/24/2012 5:41 PM, Fletcher Cocquyt wrote:
[...]
That's not relevant. The code is attempting to do the equivalent of ln -s archives/private/LISTNAME archives/public/LISTNAME What are the permissions on /var/lib/mailman/archives/public which is where it's trying to create the link? -- Mark Sapiro <mark@msapiro.net> The highway is for gamblers, San Francisco Bay Area, California better use your sense - B. Dylan
![](https://secure.gravatar.com/avatar/160625f937fa3658100cc5bce57f608d.jpg?s=120&d=mm&r=g)
Fletcher Cocquyt <fcocquyt <at> stanford.edu> writes:
I added a debug line to Archiver.py:
syslog('error', 'Archive file access failure:\n\t%s %s', old, new)
and it revealed the issue was a missing symlink in /var/lib/mailman/archives /public:
Once I corrected the ownership - it succeeded:
ls -lad /var/lib/mailman/archives/public drwxrwsr-x 2 35 utempter 32768 May 24 19:35 /var/lib/mailman/archives/public
chown mailman:mailman /var/lib/mailman/archives/public ls -lad /var/lib/mailman/archives/public drwxrwsr-x 2 mailman mailman 32768 May 24 19:35 /var/lib/mailman/archives/public
thanks
![](https://secure.gravatar.com/avatar/c5d3389a76bd697b7957049d4616a035.jpg?s=120&d=mm&r=g)
Hello All,
I am new to your the Mailman Users' group. I manage Mailman 2.1.9 (a VM Red Hat Enterprise 5 system) for my organization but I am new to Mailman and don't know much about Python scripting.
I need to generate a monthly report on all my mailing lists that includes each list's name, description, admin, and members. I have come up with a bash script using some Python scripts for Mailman that generates most of the reporting but could not find a script or a way to extract a list's description.
Any hint/help for pointing out how to extract the description of any list via a bash script will be very much appreciated.
Thanks very much for your input.
Mehdi
![](https://secure.gravatar.com/avatar/56f108518d7ee2544412cc80978e3182.jpg?s=120&d=mm&r=g)
Mehdi Mortezai wrote:
Mailman's bin/list_lists produces a list of all the installations list names and descriptions.
Any hint/help for pointing out how to extract the description of any list via a bash script will be very much appreciated.
#!/bin/bash
# assume $1 is the list name
/usr/lib/mailman/bin/list_lists | grep -i " $1 " |
sed -r "s/\s*\S+ - //"
-- Mark Sapiro <mark@msapiro.net> The highway is for gamblers, San Francisco Bay Area, California better use your sense - B. Dylan
![](https://secure.gravatar.com/avatar/c5d3389a76bd697b7957049d4616a035.jpg?s=120&d=mm&r=g)
Thanks much for the code, Mark!
Mehdi
-----Original Message----- From: Mark Sapiro [mailto:mark@msapiro.net] Sent: Friday, June 01, 2012 11:20 PM To: Mehdi Mortezai; mailman-users@python.org Subject: Re: [Mailman-Users] Help: how to extract a list's description using abash or Python script (Mailman version 2.1.9)
Mehdi Mortezai wrote:
Mailman's bin/list_lists produces a list of all the installations list names and descriptions.
Any hint/help for pointing out how to extract the description of any list via a bash script will be very much appreciated.
#!/bin/bash # assume $1 is the list name /usr/lib/mailman/bin/list_lists | grep -i " $1 " |\ sed -r "s/\s*\S+ - //"
-- Mark Sapiro <mark@msapiro.net> The highway is for gamblers, San Francisco Bay Area, California better use your sense - B. Dylan
![](https://secure.gravatar.com/avatar/56f108518d7ee2544412cc80978e3182.jpg?s=120&d=mm&r=g)
On 5/24/2012 5:41 PM, Fletcher Cocquyt wrote:
[...]
That's not relevant. The code is attempting to do the equivalent of ln -s archives/private/LISTNAME archives/public/LISTNAME What are the permissions on /var/lib/mailman/archives/public which is where it's trying to create the link? -- Mark Sapiro <mark@msapiro.net> The highway is for gamblers, San Francisco Bay Area, California better use your sense - B. Dylan
![](https://secure.gravatar.com/avatar/160625f937fa3658100cc5bce57f608d.jpg?s=120&d=mm&r=g)
Fletcher Cocquyt <fcocquyt <at> stanford.edu> writes:
I added a debug line to Archiver.py:
syslog('error', 'Archive file access failure:\n\t%s %s', old, new)
and it revealed the issue was a missing symlink in /var/lib/mailman/archives /public:
Once I corrected the ownership - it succeeded:
ls -lad /var/lib/mailman/archives/public drwxrwsr-x 2 35 utempter 32768 May 24 19:35 /var/lib/mailman/archives/public
chown mailman:mailman /var/lib/mailman/archives/public ls -lad /var/lib/mailman/archives/public drwxrwsr-x 2 mailman mailman 32768 May 24 19:35 /var/lib/mailman/archives/public
thanks
![](https://secure.gravatar.com/avatar/c5d3389a76bd697b7957049d4616a035.jpg?s=120&d=mm&r=g)
Hello All,
I am new to your the Mailman Users' group. I manage Mailman 2.1.9 (a VM Red Hat Enterprise 5 system) for my organization but I am new to Mailman and don't know much about Python scripting.
I need to generate a monthly report on all my mailing lists that includes each list's name, description, admin, and members. I have come up with a bash script using some Python scripts for Mailman that generates most of the reporting but could not find a script or a way to extract a list's description.
Any hint/help for pointing out how to extract the description of any list via a bash script will be very much appreciated.
Thanks very much for your input.
Mehdi
![](https://secure.gravatar.com/avatar/56f108518d7ee2544412cc80978e3182.jpg?s=120&d=mm&r=g)
Mehdi Mortezai wrote:
Mailman's bin/list_lists produces a list of all the installations list names and descriptions.
Any hint/help for pointing out how to extract the description of any list via a bash script will be very much appreciated.
#!/bin/bash
# assume $1 is the list name
/usr/lib/mailman/bin/list_lists | grep -i " $1 " |
sed -r "s/\s*\S+ - //"
-- Mark Sapiro <mark@msapiro.net> The highway is for gamblers, San Francisco Bay Area, California better use your sense - B. Dylan
![](https://secure.gravatar.com/avatar/c5d3389a76bd697b7957049d4616a035.jpg?s=120&d=mm&r=g)
Thanks much for the code, Mark!
Mehdi
-----Original Message----- From: Mark Sapiro [mailto:mark@msapiro.net] Sent: Friday, June 01, 2012 11:20 PM To: Mehdi Mortezai; mailman-users@python.org Subject: Re: [Mailman-Users] Help: how to extract a list's description using abash or Python script (Mailman version 2.1.9)
Mehdi Mortezai wrote:
Mailman's bin/list_lists produces a list of all the installations list names and descriptions.
Any hint/help for pointing out how to extract the description of any list via a bash script will be very much appreciated.
#!/bin/bash # assume $1 is the list name /usr/lib/mailman/bin/list_lists | grep -i " $1 " |\ sed -r "s/\s*\S+ - //"
-- Mark Sapiro <mark@msapiro.net> The highway is for gamblers, San Francisco Bay Area, California better use your sense - B. Dylan
participants (3)
-
Fletcher Cocquyt
-
Mark Sapiro
-
Mehdi Mortezai