Bug#675081: fix does not seem to work

[Matus Horvath]

After applying the fix:

nsw:/etc/init.d# /etc/init.d/proftpd restart
[....] Stopping ftp server: proftpdstart-stop-daemon: --retry takes timeout, or schedule list of at least two items
Try 'start-stop-daemon --help' for more information.
start-stop-daemon: --retry takes timeout, or schedule list of at least two items
Try 'start-stop-daemon --help' for more information.
 failed!
[ ok ] Starting ftp server: proftpd.

nsw:/etc/init.d# dpkg -l | grep proftpd
ii  proftpd-basic                 1.3.5~rc4-2                 i386    Versatile, virtual-hosting FTP daemon - binaries
ii  proftpd-doc                   1.3.5~rc4-2                 all    Versatile, virtual-hosting FTP daemon - documentation
nsw:/etc/init.d# dpkg -l | grep dpkg
ii  dpkg                          1.17.8                      i386    Debian package management system

The problem seems to be, that you need to specify full schedule to the --retry parameter, just "forever" is not enough.

nsw:/etc/init.d# start-stop-daemon --help
[---8<---]
Retry <schedule> is <item>|/<item>/... where <item> is one of
 -<signal-num>|[-]<signal-name>  send that signal
 <timeout>                       wait that many seconds
 forever                         repeat remainder forever
or <schedule> may be just <timeout>, meaning <signal>/<timeout>/KILL/<timeout>
[---8<---]

Correct way would be something like --retry $SIGNAL/forever/5/KILL.
That would first send signal, then repeat forever the sequence "wait 5 seconds, then send KILL".
On my machine this seems to fix the original bug.

Matus