Steve Holden wrote: > [EMAIL PROTECTED] wrote: > > > #import urllib, sys > > #pages = ['http://www.python.org', 'http://xxx'] > > #for i in pages: > > # try: > > # u = urllib.urlopen(i) > > # print u.geturl() > > # except Exception, e: > > # print >> sys.stderr, '%s: %s' % (e.__class__.__name__, e) > > will print an error if a page fails opening, rest opens fine > > > More generally you may wish to use the timeout features of TCP sockets. > These were introduced in Python 2.3, though Tim O'Malley's module > "timeoutsocket" (which was the inspiration for the 2.3 upgrade) was > available for earlier versions. > > You will need to import the socket module and then call > socket.setdefaulttimeout() to ensure that communication with > non-responsive servers results in a socket exception that you can trap. > > regards > Steve > -- > Steve Holden http://www.holdenweb.com/ > Python Web Programming http://pydish.holdenweb.com/ > Holden Web LLC +1 703 861 4237 +1 800 494 3119
Thank you [EMAIL PROTECTED] and Steve for some ideas.Finding the fact that the script hanged is not a big problem . I,however, would need a solution that I will not need to start again the script but the script re-start by itself. I am thinking about two threads, the main(master) that will supervise a slave thread.This slave thread will download the pages and whenever there is a timeout the master thread restart a slave thread. Is it a good solution? Or is there a better one? Thanks for help Lad -- http://mail.python.org/mailman/listinfo/python-list
