Ntp Client In Python

I've written a ntp client in python to query a time server and display the time and the program executes but does not give me any results. I'm using python's 2.7.3 integrated deve

Solution 1:

Use ntplib:

The following should work on both Python 2 and 3:

import ntplib
from time import ctime
c = ntplib.NTPClient()
response = c.request('pool.ntp.org')
print(ctime(response.tx_time))

Output:

FriJul2801:30:532017

Solution 2:

Here is a fix for the above solution, which adds fractions of seconds to the implementation and closes the socket properly. As it's actually just a handful lines of code, I didn't want to add another dependency to my project, though ntplib admittedly is probably the way to go in most cases.

#!/usr/bin/env pythonfrom contextlib import closing
from socket import socket, AF_INET, SOCK_DGRAM
import struct
import time

NTP_PACKET_FORMAT = "!12I"
NTP_DELTA = 2208988800# 1970-01-01 00:00:00
NTP_QUERY = b'\x1b' + 47 * b'\0'defntp_time(host="pool.ntp.org", port=123):
    with closing(socket( AF_INET, SOCK_DGRAM)) as s:
        s.sendto(NTP_QUERY, (host, port))
        msg, address = s.recvfrom(1024)
    unpacked = struct.unpack(NTP_PACKET_FORMAT,
                   msg[0:struct.calcsize(NTP_PACKET_FORMAT)])
    return unpacked[10] + float(unpacked[11]) / 2**32 - NTP_DELTA


if __name__ == "__main__":
    print time.ctime(ntp_time()).replace("  ", " ")

Solution 3:

It should be

msg = '\x1b' + 47 * '\0'

Instead of

msg = 'time'

But as Maksym said you should use ntplib instead.

Solution 4:

msg = '\x1b' + 47 * '\0'
.......
t = struct.unpack( "!12I", msg )[10]

Solution 5:

Sorry if my answer doesn't satisfy your expectations. I think it makes sense to use an existing solution. ntplib is a quite good library for working with NTP servers.