Python socket.error:[Errno 104]由peer重置连接

Question

我有一个代码,其中有13个客户端必须连接到服务器.然后服务器对客户端给出的数据进行一些计数.之后角色转变 - 服务器成为客户端,客户端成为接收数据的服务器.问题在于,当尝试进行第一次连接时,即13个客户端尝试连接到服务器时,我不断收到此错误: [Errno 104] Connection reset by peer.我尝试了一些解决方法,例如尝试在第二个间隔中连接5次但没有任何效果.

这是我的代码:

server.py

import socket, pickle, numpy as np
import struct
import math


while 1:
    HOST = ''
    PORT = 50007
    s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    s.bind((HOST, PORT))
    s.listen(13)

adresses = []
ports = []

i = 0
print("receiving...")
while i < 13:
    i += 1    
    #wait to accept a connection - blocking call
    conn, addr = s.accept()
    print ('Connected with ', addr)
    adresses.append(addr[0])
    buf = b''
    while len(buf) < 4:
        buf += conn.recv(4 - len(buf))
    length = struct.unpack('>I', buf)[0]
    data = b''
    l = length

    while l > 0:

        d = conn.recv(l)
        l -= len(d)
        data += d

    if not data: break

    M = np.loads(data)

    if i == 1:
        L = M[0]
    else:
        L += M[0]
    ports.append(M[1])
    conn.close() 
s.close()


L /= 993040

packet = pickle.dumps(L)
length = struct.pack('>I', len(packet))
packet = length + packet

print("sending...")
for kl, addr in enumerate(adresses):
    HOST = addr
    PORT = 50007 + ports[kl]
    s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    s.connect((HOST, PORT))
    s.sendall(packet)
    s.close()

client.py

def connection(centers, kl):

    HOST = "192.168.143.XX"
    PORT = 50007
    s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    s.settimeout(3600)
    try:
        s.connect((HOST, PORT)) # HERE IS AN ERROR
        s.settimeout(None)
        packet = pickle.dumps([centers, kl]) ## ???
        length = struct.pack('>I', len(packet))
        packet = length + packet
        s.sendall(packet) # OR HERE IS AN ERROR
        s.close()
    except Exception as e:
        print(e)
        print('error ', kl)
        s.close()
        return np.zeros(centers.shape)    


    HOST = ''
    PORT = 50007 + kl
    s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    s.bind((HOST, PORT))

    s.listen(2)

    i = 0
    while i < 1:
        #wait to accept a connection - blocking call
        conn, addr = s.accept()
        i += 1
        print ('Connected with ', addr)
        buf = b''
        while len(buf) < 4:
            buf += conn.recv(4 - len(buf))

        length = struct.unpack('>I', buf)[0]
        data = b''
        l = length
        while l > 0:
            d = conn.recv(l)
            l -= len(d)
            data += d
        if not data: break

        new_centers = np.loads(data)
        conn.close()

    s.close()
    return new_centers

aa = 0
for k in range(99):
    print(k)
    centers = some_function(centers)
    time.sleep(60)
    centers1 = connection(centers, i)
    aa = 0
    while not (centers1.any()) and aa < 5:
        time.sleep(1)
        centers1 = connection(centers, i)
        aa += 1
    centers = centers1

事情是所有13个客户端都必须连接到服务器,否则它将不会进行下一次迭代.我正在使用Python 3.4.请帮忙.

更新:

我添加了线程,但错误仍然存​​在:

[Errno 104]通过对等方重置连接

server.py

import socket, pickle, numpy as np
import struct
import math
from multiprocessing.pool import ThreadPool

def clientthread(conn, L):
    buf = b''
    while len(buf) < 4:
        buf += conn.recv(4 - len(buf))
    length = struct.unpack('>I', buf)[0]
    data = b''
    l = length
    while l > 0:
        d = conn.recv(l)
        l -= len(d)
        data += d
    M = np.loads(data)


    return(M)


j = 0
while 1:
    HOST = ''
    PORT = 50007
    s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    s.bind((HOST, PORT))
    s.listen(2)
    #print('0')

    adresses = []
    ports = []

    i = 0
    print("receiving...")
    while i < 13:
        i += 1    
        #wait to accept a connection - blocking call
        conn, addr = s.accept()
        print ('Connected with ', addr)
        adresses.append(addr[0])
        pool = ThreadPool(processes=13)
        async_result = pool.apply_async(clientthread, (conn, i,))
        M = async_result.get()  
        conn.close() 

        if i == 1:
            L = M[0]
        else:
            L += M[0]
        ports.append(M[1])
    s.close()


    L /= 993040

    packet = pickle.dumps(L)
    length = struct.pack('>I', len(packet))
    packet = length + packet

    for kl, addr in enumerate(adresses):
        HOST = addr
        PORT = 50007 + ports[kl]
        s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
        s.connect((HOST, PORT))
        s.sendall(packet)
        s.close()

Answer 1

客户端似乎已连接到服务器，但[Errno 104] Connection reset by peer在尝试发送数据时遇到了“ ”异常。Python 第一次引发“[Errno 104] Connection reset by peer”异常，然后第二次和更多你会在客户端得到“[Errno 32] Broken pipe”异常。

这可能意味着服务器已启动并正在侦听端口（否则，您会得到“ [Errno 111] Connection refused" exception on the client side”。这也意味着服务器在关闭连接之前崩溃了，因为如果在客户端发送数据之前在服务器端关闭了连接一边，客户端会遇到“ [Errno 32] Broken pipe”异常。

“由对等方重置连接”是 TCP/IP 等价物，相当于将电话重新挂断。这比仅仅不回复而留下一个悬而未决更有礼貌。但这不是真正礼貌的 TCP/IP 对话者所期望的 FIN-ACK。（来自另一个stackoverflow答案）

Answer 2

您将必须在服务器上使用线程。
有关简单的解释，请参阅：http://www.binarytides.com/python-socket-server-code-example/
给出的简单示例的内容是：

import socket
import sys
from thread import *

HOST = ''   # Symbolic name meaning all available interfaces
PORT = 8888 # Arbitrary non-privileged port

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
print 'Socket created'

#Bind socket to local host and port
try:
    s.bind((HOST, PORT))
except socket.error as msg:
    print 'Bind failed. Error Code : ' + str(msg[0]) + ' Message ' + msg[1]
    sys.exit()

print 'Socket bind complete'

#Start listening on socket
s.listen(10)
print 'Socket now listening'

#Function for handling connections. This will be used to create threads
def clientthread(conn):
    #Sending message to connected client
    conn.send('Welcome to the server. Type something and hit enter\n') #send only takes string

    #infinite loop so that function do not terminate and thread do not end.
    while True:

        #Receiving from client
        data = conn.recv(1024)
        reply = 'OK...' + data
        if not data: 
            break

        conn.sendall(reply)

    #came out of loop
    conn.close()

#now keep talking with the client
while 1:
    #wait to accept a connection - blocking call
    conn, addr = s.accept()
    print 'Connected with ' + addr[0] + ':' + str(addr[1])

    #start new thread takes 1st argument as a function name to be run, second is the tuple of arguments to the function.
    start_new_thread(clientthread ,(conn,))

s.close()