On 11/15/2011 8:48 AM, Chris Petrik wrote:
>
> % cd lists/services-officialunix.com
> %ls -l |wc -l
> 8
> %
> %ls -l
> total 24
> -rw-rw---- 1 www mailman 3792 Nov 15 07:35 config.pck
> -rw-rw---- 1 mailman mailman 3792 Nov 15 08:03
> config.pck.tmp.hosting.officialunix.com.30747
> -rw-rw---- 1 mailman mailman 3835 Nov 15 08:00
> config.pck.tmp.hosting.officialunix.com.74443
> -rw-rw---- 1 mailman mailman 3792 Nov 15 07:48
> config.pck.tmp.hosting.officialunix.com.74445
> -rw-rw---- 1 mailman mailman 3792 Nov 15 08:00
> config.pck.tmp.hosting.officialunix.com.74446
> -rw-rw---- 1 mailman mailman 131 Nov 15 07:37 pending.pck
> -rw-rw-r-- 1 mailman mailman 599 Nov 15 07:37 request.pck
> %ln config.pck.tmp.hosting.officialunix.com.74446 config.pck.last
> %ls
> config.pck
> config.pck.last
Actually, I gave you the wrong advice. The ln command should have been
ln config.pck config.pck.last
If that works, instead of trying this via the ln command, you could try
it in a python process. Again as user mailman in mailman's home directory do
bin/withlist -i
This will respond with a few lines followed by a >>> prompt. At the
prompts enter
import os
os.unlink('lists/services-officialunix.com/config.pck.last')
(ignore any non-existant file exception)
os.link('lists/services-officialunix.com/config.pck',
'lists/services-officialunix.com/config.pck.last')
You can enter that all on one line, or you can enter
os.link('lists/services-officialunix.com/config.pck',
which will result in a ... prompt to which you enter
'lists/services-officialunix.com/config.pck.last')
If that works, the question is why can you do it as mailman, but the
qrunners can't? Did you start Mailman by running bin/mailmanctl start as
root? If not, stop Mailman and then start is as root.
--
Mark Sapiro <m...@msapiro.net> The highway is for gamblers,
San Francisco Bay Area, California better use your sense - B. Dylan
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