Re: urllib2 error
On Wed, Nov 17, 2010 at 5:31 PM, Kushal Kumaran wrote: > On Wed, Nov 17, 2010 at 5:18 PM, asit wrote: >> I have this piece of code >> >> import urllib2 >> >> proc_url = 'http://www.nse-india.com/content/historical/EQUITIES/2001/ >> JAN/cm01JAN2001bhav.csv.zip' >> print 'processing', proc_url >> req = urllib2.Request(proc_url) >> res = urllib2.urlopen(req) >> >> when i run this...following error comes >> >> Traceback (most recent call last): >> >> HTTPError: HTTP Error 403: Forbidden >> > > The web server is probably trying to forbid downloads using automated > tools by examining the user agent header (downloading using wget also > fails, for example). You can work around that by setting the > User-Agent header to the same value as a browser that works. > > The urllib2 documentation covers how to set headers in requests. > > -- > regards, > kushal > -- Regards, Asit Kumar Dhal TCS, Mumbai blog: http://asitdhal.blogspot.com/ Thanx...it worked -- http://mail.python.org/mailman/listinfo/python-list
urllib2 error
I have this piece of code import urllib2 proc_url = 'http://www.nse-india.com/content/historical/EQUITIES/2001/ JAN/cm01JAN2001bhav.csv.zip' print 'processing', proc_url req = urllib2.Request(proc_url) res = urllib2.urlopen(req) when i run this...following error comes Traceback (most recent call last): File "C:/Python26/reqnse.py", line 92, in res = urllib2.urlopen(req) File "C:\Python26\lib\urllib2.py", line 124, in urlopen return _opener.open(url, data, timeout) File "C:\Python26\lib\urllib2.py", line 395, in open response = meth(req, response) File "C:\Python26\lib\urllib2.py", line 508, in http_response 'http', request, response, code, msg, hdrs) File "C:\Python26\lib\urllib2.py", line 433, in error return self._call_chain(*args) File "C:\Python26\lib\urllib2.py", line 367, in _call_chain result = func(*args) File "C:\Python26\lib\urllib2.py", line 516, in http_error_default raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) HTTPError: HTTP Error 403: Forbidden would anyone help me why ?? -- http://mail.python.org/mailman/listinfo/python-list
indentation error
Consider the following code
import stat, sys, os, string, commands
try:
pattern = raw_input("Enter the file pattern to search for :\n")
commandString = "find " + pattern
commandOutput = commands.getoutput(commandString)
findResults = string.split(commandOutput, "\n")
print "Files : "
print commandOutput
print "="
for file in findResults:
mode = stat.S_IMODE(os.lstat(file)[stat.ST_MODE])
print "\nPermissions for file", file, ":"
for level in "USR", "GRP", "OTH":
for perm in "R", "W", "X":
if mode & getattr(stat,"S_I"+perm+level):
print level, " has ", perm, " permission"
else:
print level, " does NOT have ", perm, "
permission"
except:
print "There was a problem - check the message above"
According to me, indentation is ok. but the python interpreter gives
an indentation error
[asit ~/py] $ python search.py
File "search.py", line 7
findResults = string.split(commandOutput, "\n")
^
IndentationError: unindent does not match any outer indentation level
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Re: process mp3 file
> Define "processing". getting the title, song name, etc of the file and updating in a database -- http://mail.python.org/mailman/listinfo/python-list
process mp3 file
Somebody suggest me a python library for processing mp3 file. Here I don't want to play the file. Thank you -- http://mail.python.org/mailman/listinfo/python-list
fetch all tweets..
This question is for any python-twitter developer
I want to develop an application using python twitter .
Just look at the code...
import twitter
api = twitter.Api();
sta = api.GetUserTimeline('ShashiTharoor')
i = 0
for s in sta:
i +=1
print str(i) + " " + s.text
print
print
The above code only fetches 20 tweets. How can I fetch all tweets ??
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database handling
I need some tutorial about python-mysql connectivity(database handling). Somebody please help me !! -- http://mail.python.org/mailman/listinfo/python-list
Re: exception due to NoneType
On Nov 7, 10:36 pm, Bruno Desthuilliers
wrote:
> asit a écrit :
>
> > In my program I want to catch exception which is caused by accessing
> > NoneType object.
>
> > Can anyone suggest me how this can be done ??
>
> Not without the minimal working code exposing your problem, or the full
> traceback you got. Merely "accessing NoneType object" doesn't by itself
> raise any exception... I suspect you get the None object where you
> expected something else and try to access an attribute of this
> 'something else', and ends up getting an AttributeError, but there are
> other possible scenarii that might fit your (very poor) description of
> the problem, so no way too help you without more informations. As a
> general rule, remember that the traceback is actually meant to *help*
> finding out what went wring.
I could have described the error, but the problem is that it's
dependent of a third party library..
Let me write the code here...
import twitter
api = twitter.Api('asitdhal','swordfish')
users = api.GetFriends()
for s in users:
print
print "##"
try:
print "user id : " + str(s.id)
print "user name : " + s.name
print "user location : " + s.location
print "user description : " + s.description
print "user profile image url : " + s.profile_image_url
print "user url : " + s.url
print "user status : " + str(s.status)
except TypeError:
pass
look at the except TypeError. This is supposed to catch only exception
thrown by NoneType.
please help me.
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exception due to NoneType
In my program I want to catch exception which is caused by accessing NoneType object. Can anyone suggest me how this can be done ?? -- http://mail.python.org/mailman/listinfo/python-list
exception in urllib2
I hv been developing a link scanner. Here the objective is to
recursively scan a particular web site.
During this, my script met http://images.google.co.in/imghp?hl=en&tab=wi
and passed it to the scan function, whose body is like this..
def scan(site):
log=open(logfile,'a')
log.write(site + "\n")
site = "http://"; + site.lower()
try:
site_data = urllib.urlopen(site)
parser = MyParser()
parser.parse(site_data.read())
except(IOError),msg:
print "Error in connecting site ", site
print msg
links = parser.get_hyperlinks()
for l in links:
log.write(l + "\n")
But it throws a weird exception like this...
Traceback (most recent call last):
File "I:\Python26\linkscan1.py", line 104, in
main()
File "I:\Python26\linkscan1.py", line 95, in main
scan(lk)
File "I:\Python26\linkscan1.py", line 65, in scan
site_data = urllib.urlopen(site)
File "I:\Python26\lib\urllib.py", line 87, in urlopen
return opener.open(url)
File "I:\Python26\lib\urllib.py", line 203, in open
return getattr(self, name)(url)
File "I:\Python26\lib\urllib.py", line 327, in open_http
h = httplib.HTTP(host)
File "I:\Python26\lib\httplib.py", line 984, in __init__
self._setup(self._connection_class(host, port, strict))
File "I:\Python26\lib\httplib.py", line 656, in __init__
self._set_hostport(host, port)
File "I:\Python26\lib\httplib.py", line 668, in _set_hostport
raise InvalidURL("nonnumeric port: '%s'" % host[i+1:])
httplib.InvalidURL: nonnumeric port: ''
How can i handle this ???
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Re: Handle SystemExit exception
Thanx everyone This is my fault. Exception was thrown before the main function. -- http://mail.python.org/mailman/listinfo/python-list
Handle SystemExit exception
My program contains some sys.exit(1) and when its executed an exception known as SystemExit is thrown. I tried to handle bu using following code snippet if __name__ == "__main__": try: main() #main contains sys.exit(1) except KeyboardInterrupt: print "aborted by user" except SystemExit: pass But it does not work. Can anyone help me out ??? Though KeyboradInterrupt is handled, but why SystemExit is not handles Regards Asit Dhal -- http://mail.python.org/mailman/listinfo/python-list
process command line parameter
Recently I was coding a link extractor. It's a command line stuff and takes parameter as argument. I found that the in operator is not always helpful. eg. if "--all" in sys.argv: print "all links will be printed" its not helpful when some attribute value is sent in command line parameter. hence I want to process some data like find=".co.uk" How can i do this ??? -- http://mail.python.org/mailman/listinfo/python-list
Re: why o/p is different ???
On Jan 15, 11:47 am, "James Mills" wrote: > On Thu, Jan 15, 2009 at 4:34 PM, asit wrote: > > I recently faced a peculiar o/p. > > > My objective is to remove the command name(my script name) from > > sys.argv[0]. > > I coded like this > > If you _really_ want to remove your script_name from > sys.argv, then do this: > > del sys.argv[0] > > If you're just after what the name of your script is > that's being run (for human readability) try the > following function (taken from pymills): > > def getProgName(): > """getProgName() -> str > > Return the name of the current program being run > by working it out from the script's basename. > """ > > return os.path.basename(sys.argv[0]) > > -- > > cheers > James Thank you everyone -- http://mail.python.org/mailman/listinfo/python-list
why o/p is different ???
I recently faced a peculiar o/p.
My objective is to remove the command name(my script name) from
sys.argv[0].
I coded like this
import urllib
import sys
print "\n\n\t\tlipun4u[at]gmail[dot]com"
print "\t\t"
apppath = sys.argv[0].split("/")
appname = apppath[len(apppath)-1]
print appname
if len(sys.argv) not in [2,3]:
print "Usage : " + appname + " [options]"
print "e.g. : " + appname + "www.google.com --verbose"
print "\n\t[option]"
print "\t\t--verbose/-V for verbose output"
print "\t\t-r for recursive scan"
sys.exit(1)
site = appname.replace("http://","";).rsplit("/",1)[0]
site = "http://"; + site.lower()
print site
It showed the required o/p in the IDLE
here is the o/p
lipun4u[at]gmail[dot]com
linkscan.py
Usage : linkscan.py [options]
e.g. : linkscan.pywww.google.com --verbose
[option]
--verbose/-V for verbose output
-r for recursive scan
Traceback (most recent call last):
File "I:/Python26/linkscan.py", line 18, in
sys.exit(1)
SystemExit: 1
But in command prompt the o/p is still faulty
lipun4u[at]gmail[dot]com
I:\Python26\linkscan.py
Usage : I:\Python26\linkscan.py [options]
e.g. : I:\Python26\linkscan.pywww.google.com --verbose
[option]
--verbose/-V for verbose output
-r for recursive scan
I:\Python26>
regards
asit dhal
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fetch image
import httplib
class Server:
#server class
def __init__(self, host):
self.host = host
def fetch(self, path):
http = httplib.HTTPConnection(self.host)
http.request("GET", path)
r = http.getresponse()
print str(r.status) + " : " + r.reason
server = Server("www.python.org")
fp=open("phpvuln.txt")
x=fp.readlines();
for y in x:
server.fetch("/" + y);
The above code fetches only the html source of the webpage. How to get
the image, flash animation and other stuffs
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Re: ResponseNotReady exception
On Jan 4, 1:59 am, "Chris Rebert" wrote:
> On Sat, Jan 3, 2009 at 12:38 PM, asit wrote:
> > import httplib
>
> > class Server:
> > #server class
> > def __init__(self, host):
> > self.host = host
> > def fetch(self, path):
> > http = httplib.HTTPConnection(self.host)
> > http.putrequest("GET", path)
>
> According to the docs, if you use .putrequest(), you have to also then
> call .putheader() [as many times as necessary], .endheaders(), and
> .send(), in that order. I'd advise just changing .putrequest() to
> .request() instead.
>
> And you should also read the fine docs
> yourself:http://docs.python.org/library/httplib.html
>
> Cheers,
> Chris
>
> --
> Follow the path of the Iguana...http://rebertia.com
Thanx
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ResponseNotReady exception
import httplib
class Server:
#server class
def __init__(self, host):
self.host = host
def fetch(self, path):
http = httplib.HTTPConnection(self.host)
http.putrequest("GET", path)
r = http.getresponse()
print str(r.status) + " : " + r.reason
server = Server("www.python.org")
fp=open("phpvuln.txt")
x=fp.readlines();
for y in x:
server.fetch("/" + y);
In the above code the phpvuln.txt file contains a list of directories.
Why the above code throws ResponseNotReady exception ???
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Re: best IDE
On Nov 26, 11:09 pm, "Chris Rebert" <[EMAIL PROTECTED]> wrote: > On Wed, Nov 26, 2008 at 9:59 AM, asit <[EMAIL PROTECTED]> wrote: > > Which one is the best IDE for python > > This was recently discussed. To avoid needlessly rehashing said > discussion, see the thread > athttp://groups.google.com/group/comp.lang.python/browse_thread/thread/... > > Cheers, > Chris > -- > Follow the path of the Iguana...http://rebertia.com > > > -- > >http://mail.python.org/mailman/listinfo/python-list Thanx -- http://mail.python.org/mailman/listinfo/python-list
best IDE
Which one is the best IDE for python -- http://mail.python.org/mailman/listinfo/python-list
Re: compare items in list to x
On Nov 2, 9:54 pm, "Pete Kirkham" <[EMAIL PROTECTED]> wrote: > 2 is not equal to '2' As the error is correctly marked, u have to convert '2' to 2 by using int() function so the write code is x = 2 def compare(): for y in ['12', '33', '2']: y=int(y) #string '2' is converted to int 2 if x < y: print x, "is less than", y elif x > y: print x, "is greater than", y else: print x, "and", y, "are equal" compare() Regards Asit Dhal -- http://mail.python.org/mailman/listinfo/python-list
Re: project in python
On Oct 28, 10:02 am, alex23 <[EMAIL PROTECTED]> wrote: > On Oct 26, 2:51 am, Stefan Behnel <[EMAIL PROTECTED]> wrote: > > > The more you spam people with your repetitive postings, the less likely it > > becomes that they are willing to answer you. > > In asit's defence, the Google Groups interface has been woefully > broken for the past 3-4 days. If e had posted via it, messages > wouldn't have been visible to em until today at the earliest. After all these search and queries, I have decided to make smething about Google API. Is there an API available for google group ??? -- http://mail.python.org/mailman/listinfo/python-list
XML-RPC
what is XML-RPC System -- http://mail.python.org/mailman/listinfo/python-list
project in python
I am a newbie and learned python to some extent. I want to do some project in python based on network programming or HTML/XML parsing. Can anyone suggest me about this ??? -- http://mail.python.org/mailman/listinfo/python-list
project in python
I want to do a project in python. It should be something based on socket programming, HTML/XML parsing, etc plz suggest me -- http://mail.python.org/mailman/listinfo/python-list
project in python
I want to do a project in python. It should be something based on socket programming, HTML/XML parsing, etc please suggest me -- http://mail.python.org/mailman/listinfo/python-list
Re: portable python
On Oct 24, 11:18 pm, "Jerry Hill" <[EMAIL PROTECTED]> wrote: > On Fri, Oct 24, 2008 at 1:42 PM, asit <[EMAIL PROTECTED]> wrote: > > I code in both windows and Linux. As python is portable, the o/p > > should be same in both cases. But why the following code is perfect in > > windows but error one in Linux ??? > > What error message do you get in linux? How are you running your code > in linux? Your code seems to generally work on my Ubuntu linux box, > so you need to give us more information. > > -- > Jerry this the o/p [EMAIL PROTECTED]:~/hack$ python portscan.py 59.93.128.10 10 20 Traceback (most recent call last): File "portscan.py", line 33, in print str(port) + " : " + scan(ip,port,timeout) File "portscan.py", line 22, in scan return status[result] KeyError: 11 [EMAIL PROTECTED]:~/hack$ -- http://mail.python.org/mailman/listinfo/python-list
portable python
I code in both windows and Linux. As python is portable, the o/p
should be same in both cases. But why the following code is perfect in
windows but error one in Linux ???
from socket import *
import sys
status={0:"open",10049:"address not available",10061:"closed",
10060:"timeout",10056:"already connected",10035:"filtered",11001:"IP
not found",10013:"permission denied"}
def scan(ip,port,timeout):
s = socket(AF_INET, SOCK_STREAM)
s.settimeout(timeout)
try:
result= s.connect_ex((ip, port))
except:
print "Cannot connect to IP"
return
s.close()
return status[result]
if (len(sys.argv) == 4):
ip=sys.argv[1]
minrange = int(sys.argv[2])
maxrange = int(sys.argv[3])
timeout = 3
ports=range(minrange,maxrange+1)
for port in ports:
print str(port) + " : " + scan(ip,port,timeout)
else:
print "usage : " + sys.argv[0] + " "
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Re: dictionary
On Oct 24, 8:01 pm, Steven D'Aprano <[EMAIL PROTECTED] cybersource.com.au> wrote: > On Fri, 24 Oct 2008 14:53:19 +, Peter Pearson wrote: > > On 24 Oct 2008 13:17:45 GMT, Steven D'Aprano wrote: > > >> What are programmers coming to these days? When I was their age, we > >> were expected to *read* the error messages our compilers gave us, not > >> turn to the Interwebs for help as soon there was the tiniest problem. > > > Yes, and what's more, the text of the error message was "IEH208". After > > reading it several times, one looked it up in a big fat set of books, > > where one found the explanation: > > > IEH208: Your program contains an error. Correct the error and resubmit > > your job. > > > An excellent system for purging the world of the weak and timid. > > You had reference books? You were lucky! When I was lad, we couldn't > afford reference books. If we wanted to know what an error code meant, we > had to rummage through the bins outside of compiler vendors' offices > looking for discarded documentation. > > -- > Steven I don't have a reference book. I read from e-buks and some print outs of Reference Card. Again, I had this error becoz my console has no colour highlighting feature. -- http://mail.python.org/mailman/listinfo/python-list
Re: dictionary
On Oct 24, 3:06 pm, Tim Chase <[EMAIL PROTECTED]> wrote: > ["%s="%s" % (k,v) for k,v in d.items()] > > File "", line 1 > > ["%s="%s" % (k,v) for k,v in d.items()] > > ^ > > SyntaxError: EOL while scanning single-quoted string > > You have three quotation marks... you want > >"%s=%s" > > not > >"%s="%s" > > -tkc Thanx -- http://mail.python.org/mailman/listinfo/python-list
dictionary
what the wrong with the following code
>>> d={"server":"mpilgrim","database":"master",
... "uid":"sa",
... "pwd":"secret"}
>>> d
{'pwd': 'secret', 'database': 'master', 'uid': 'sa', 'server':
'mpilgrim'}
>>> ["%s="%s" % (k,v) for k,v in d.items()]
File "", line 1
["%s="%s" % (k,v) for k,v in d.items()]
^
SyntaxError: EOL while scanning single-quoted string
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Re: mulithreaded server
On Mar 11, 9:10 pm, Jean-Paul Calderone <[EMAIL PROTECTED]> wrote:
> On Tue, 11 Mar 2008 08:24:54 -0700 (PDT), asit <[EMAIL PROTECTED]> wrote:
> >import socket
> >import sys
> >import thread
>
> >p=1
> >PORT=11000
> >BUFSIZE=1024
>
> >def getData(cSocket):
> >global stdoutlock,cSocketlock
> >while True:
> >cSocketlock.acquire()
> >data=cSocket.recv(BUFSIZE)
> >if data=='q':
> >data='client exited'
> >cSocket.close()
> >p=0
> >cSocketlock.release()
> >stdoutlock.acquire()
> >stdout.write(data)
> >stdoutlock.release()
>
> >def sendData(cSocket):
> >global stdoutlock,cSocketlock
> >while True:
> >stdoutlock.acquire()
> >data=raw_input('>>')
> >cSocketlock.acquire_lock()
> >if data=='q':
> >stdout.write('server exited')
> >stdout.release()
> >p=0
> >cSocket.close()
> >sSocket.send(data)
> >sSocketlock.release()
>
> Could it be because `sSocketlock´ here
>
>
>
> >stdout=sys.stdout
> >host=''
> >sSocket=socket.socket(socket.AF_INET,socket.SOCK_STREAM)
> >sSocket.bind((host,PORT,))
> >sSocket.listen(1)
> >#sSocketlock=thread.allocate_lock()
>
> is never bound since the line above here is commented out?
>
> >stdoutlock=thread.allocate_lock()
> >print 'waiting for connection'
> >cSocket,addr=sSocket.accept()
> >print 'connection from',addr
> >cSocketlock=thread.allocate_lock()
> >thread.start_new_thread(sendData,(cSocket,))
> >thread.start_new_thread(getData,(cSocket,))
> >if p==0:
> >sSocket.close()
>
> >In the above program, why there is an unhandeled exception ???
>
> Just a guess. You should really include the traceback when you ask a
> question like this.
>
> Jean-Paul
It's not a traceback error. It's an unhandeled exception..please help
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mulithreaded server
import socket
import sys
import thread
p=1
PORT=11000
BUFSIZE=1024
def getData(cSocket):
global stdoutlock,cSocketlock
while True:
cSocketlock.acquire()
data=cSocket.recv(BUFSIZE)
if data=='q':
data='client exited'
cSocket.close()
p=0
cSocketlock.release()
stdoutlock.acquire()
stdout.write(data)
stdoutlock.release()
def sendData(cSocket):
global stdoutlock,cSocketlock
while True:
stdoutlock.acquire()
data=raw_input('>>')
cSocketlock.acquire_lock()
if data=='q':
stdout.write('server exited')
stdout.release()
p=0
cSocket.close()
sSocket.send(data)
sSocketlock.release()
stdout=sys.stdout
host=''
sSocket=socket.socket(socket.AF_INET,socket.SOCK_STREAM)
sSocket.bind((host,PORT,))
sSocket.listen(1)
#sSocketlock=thread.allocate_lock()
stdoutlock=thread.allocate_lock()
print 'waiting for connection'
cSocket,addr=sSocket.accept()
print 'connection from',addr
cSocketlock=thread.allocate_lock()
thread.start_new_thread(sendData,(cSocket,))
thread.start_new_thread(getData,(cSocket,))
if p==0:
sSocket.close()
In the above program, why there is an unhandeled exception ???
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Re: message entry box at center
On Feb 28, 7:53 pm, Peter Otten <[EMAIL PROTECTED]> wrote: > asit wrote: > > i want to show the entry button at the center of the window. How is it > > possible ?? > > from Tkinter import * > > > def callback(): > > print e.get() > > > master=Tk() > > e=Entry(master) > > e.pack(expand=True) > > > e.focus_set() > > > b=Button(master,text="get",width=10,command=callback) > > b.pack(anchor=CENTER) > > > master.mainloop() > > For more complex layouts have a look at the grid geometry manager. > > Peter but here there is another problem. there is a huge gap between get button and message entry button when maximized. pl z help. !! -- http://mail.python.org/mailman/listinfo/python-list
message entry box at center
from Tkinter import * def callback(): print e.get() master=Tk() e=Entry(master) e.pack(anchor=CENTER) e.focus_set() b=Button(master,text="get",width=10,command=callback) b.pack(anchor=CENTER) master.mainloop() i want to show the entry button at the center of the window. How is it possible ?? -- http://mail.python.org/mailman/listinfo/python-list
class object using widget
from Tkinter import * # get widget classes
from tkMessageBox import askokcancel # get canned std dialog
class Quitter(Frame): # subclass our GUI
def __init__(self, parent=None): # constructor method
Frame.__init__(self, parent)
self.pack()
widget = Button(self, text='Quit', command=self.quit)
widget.pack(side=LEFT)
def quit(self):
ans = askokcancel('Verify exit', "Really quit?")
if ans: Frame.quit(self)
if __name__ == '__main__': Quitter().mainloop()
In the above program, why the error comes ??
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inode number in windows XP
why this program shows ambiguous behavior ??
import os
import stat
import time
#import types
file_name=raw_input("Enter file name : ")
print file_name, "information"
st=os.stat(file_name)
print "mode", "=>", oct(stat.S_IMODE(st[stat.ST_MODE]))
print "type","=>",
if stat.S_ISDIR(st[stat.ST_MODE]):
print "DIReCTORY"
elif stat.S_ISREG(st[stat.ST_MODE]):
print "REGULAR"
elif stat.S_ISLINK(st[stat.ST_MODE]):
print "LINK"
print "file size", "=>",st[stat.ST_SIZE]
print "inode number", "=>",st[stat.ST_INO]
print "device inode resides on", "=>",st[stat.ST_DEV]
print "number of links to this inode", "=>",st[stat.ST_NLINK]
print "last accessed", "=>", time.ctime(st[stat.ST_ATIME])
print "last modified", "=>", time.ctime(st[stat.ST_MTIME])
print "inode changed", "=>", time.ctime(st[stat.ST_CTIME])
i ran this program in Winows XP SP2 in python 2.5.
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time.gmtime
we know that time.gmtime(secs) takes a parameter secs. what does this secs suggest ??What is it's significance ?? -- http://mail.python.org/mailman/listinfo/python-list
