(Re: Debian bug report #479346 - http://bugs.debian.org/cgi-bin/bugreport.cgi?bug=479346) (Copying over to the bug report itself, please send copies of your replies as well)
The cherokee-admin program is a nice and nifty tool... But it lacks
some polish which can easily scare away users. This bug report is a
very good example: cherokee-admin launches a statically configured
Cherokee server, running the admin program via a SCGI
connection... Statically configured to be at TCP port 4000:
cherokee/admin_main.c line 74:
RULE "1!handler!balancer!local1!host = localhost:4000\n"
Granted, we cannot expect much configurability from the configurator,
right? :-) The choice of port 4000 is IMHO unfortunate, as it is a
common port for local use. And it is not documented at all. But the
real problem is the conflict is not reported - The user just gets a
«500 Internal Server Error» message, but nothing in the console.
I just talked with Álvaro, and he agrees that it's a good (and simple
to implement) idea for cherokee-admin to check if the port is
available upon invocation, and send a prominent message to the user in
the console, refusing to run. But in any case, a more robust method
would be very nice to have: Would any of you feel like coding a
failover mechanism, that looks for a free port? I am completely
unfamiliar with the mechanism, so I'd rather ask for a patch than
write one myself ;-)
In the meantime, I'm modifying the Debian package, so instead of
binding to such a popular port (at least it conflicts with mldonkey)
it binds to something else... say, port 31337? No, I guess that's
popular as well ;-) Well, a random high number. Of course, I'll check
it.
And, just for extra points: When you want to close the cherokee-admin
session, you kill the process with an innocuous Ctrl-C, right? Well,
the only problem is it looks as if we hit an error:
$$ /usr/sbin/cherokee-admin
Cherokee Web Server 0.6.1 (May 5 2008): Listening on port 9090, TLS
disabled
IPv6 disabled, using epoll, 1024 fds system limit, max. 502 connections
10 threads, 105 fds per thread, standard scheduling policy
Server running.. PID=7355
Traceback (most recent call last):
File "/usr/share/cherokee/admin/server.py", line 164, in <module>
main()
File "/usr/share/cherokee/admin/server.py", line 158, in main
srv.handle_request()
File "/usr/lib/python2.5/SocketServer.py", line 217, in handle_request
request, client_address = self.get_request()
File "/usr/lib/python2.5/SocketServer.py", line 374, in get_request
return self.socket.accept()
File "/usr/lib/python2.5/socket.py", line 167, in accept
sock, addr = self._sock.accept()
KeyboardInterrupt
So, of course, this is just matter of anticipating we will be
terminated by a Ctrl-C, avoiding a head-scratching user ;-) So, would
anybody object to this? (I'm submitting it to the list as I'm a
complete python newbie, and I don't grok cherokee internals)
Index: cherokee-0.7.1/admin/server.py
===================================================================
--- cherokee-0.7.1.orig/admin/server.py 2008-06-11 14:54:46.000000000 -0500
+++ cherokee-0.7.1/admin/server.py 2008-06-11 14:55:56.000000000 -0500
@@ -158,11 +158,14 @@
cfg = Config(cfg_file)
print ("Server running.. PID=%d" % (os.getpid()))
- while True:
- # Do it
- srv.handle_request()
+ try:
+ while True:
+ # Do it
+ srv.handle_request()
- srv.server_close()
+ srv.server_close()
+ except KeyboardInterrupt:
+ print "Server exiting..."
if __name__ == '__main__':
main()
I hope it makes sense (I have not even tested it yet ;-) )
Thank you,
--
Gunnar Wolf - [EMAIL PROTECTED] - (+52-55)5623-0154 / 1451-2244
PGP key 1024D/8BB527AF 2001-10-23
Fingerprint: 0C79 D2D1 2C4E 9CE4 5973 F800 D80E F35A 8BB5 27AF
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