2009/2/18 Tom Davis <[email protected]>:
>
> I am trying to use multiprocessing within a modwsgi app and am having
> an issue. At first I was getting errors regarding permissions to use
> stdout, etc. I fixed these by setting WSGIRestrictStdout,
> WSGIRestrictStdin, and WSGIRestrictSignal to Off in my conf file. Now
> I get no errors, but the process doesn't actually run. My usage is
> quite simple: on a page request, I try something along the lines of:
>
> p = multiprocessing.Process(target=fn, args...)
> p.start()
> p.join()
>
> The target doesn't output anything, but if run properly it should add
> some records to the database. I tested that it works normally by
> running the app under the django development server, which doesn't
> seem to have an issue with multiprocessing. Has anyone encountered
> this before?
What does the target function do?
Try following WSGI script file. Note that for this script you will
need to look in main Apache error log for output, even if WSGI script
defined in VirtualHost context and that VirtualHost has its own error
log. This is because am reseting stdin/stdout/stderr back to original
values. Take out the WSGIRestrictStdin, WSGIRestrictStdout and
WSGIRestrictSignal directives you already added.
from multiprocessing import Process
import sys, os
sys.stdin = sys.__stdin__
sys.stdout = sys.__stdout__
sys.stderr = sys.__stderr__
def f(name):
sys.stderr.write('subprocess pid %d\n' % os.getpid())
sys.stderr.write('hello %s\n' % name)
sys.stderr.flush()
def application(environ, start_response):
status = '200 OK'
output = 'Hello World!'
sys.stderr.write('parent pid %d\n' % os.getpid())
sys.stderr.flush()
p = Process(target=f, args=('grumpy',))
p.start()
p.join()
response_headers = [('Content-type', 'text/plain'),
('Content-Length', str(len(output)))]
start_response(status, response_headers)
return [output]
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