Long Ge wrote:
import socket
if __name__ == '__main__':
print("main")
udp = socket.socket(socket.AF_INET,socket.SOCK_DGRAM)
udp.sendto("1",0,('127.0.0.1 <http://127.0.0.1>',3722))
udp.close()
*output:*
main
Traceback (most recent call last):
File "C:\Python30\Projects\udp.py", line 6, in <module>
udp.sendto("1",0,('127.0.0.1 <http://127.0.0.1>',3722))
TypeError: sendto() argument 1 must be bytes or read-only buffer, not str
what is the problem about argument 1?
Python 3.0 is much stricter than 2.x about the distinction between
character based-data (i.e. Unicode strings) and 8-bit byte sequences
(bytes and bytearray).
As Benjamin pointed out, socket.sendto expects a bytestream rather than
a character string, so you need to either use bytes directly (such as
b"1") or encode the character string to a byte sequence (such as
"1".encode("utf-8")).
Cheers,
Nick.
--
Nick Coghlan | [EMAIL PROTECTED] | Brisbane, Australia
---------------------------------------------------------------
http://www.boredomandlaziness.org
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