字典更新内部功能，但不在外面

我有一个线程（new_client_thread）更新字典并打印它。当我试图再次从主线程打印字典时，字典没有改变。当线程（new_client_thread）再次更改字典时，所有更改都被保存了，但只在该线程内而不在主内。字典更新内部功能，但不在外面

这是我server.py

from threading import Thread 
from threading import Lock 


def add_client_to_dict(my_id, connect): 
    my_socket_list[int(my_id)] = connect 
    print my_socket_list 


def print_my_socket_list(): 
    print my_socket_list 
    if len(my_socket_list) == 0: 
     print "dictionary is empty" 
     return None 
    for item in my_socket_list: 
     print item 


def connection_failed(connecting): 
    print "deleting" 
    for key in my_socket_list.keys(): 
     if my_socket_list[key] == connecting: 
      my_socket_list.pop(key) 
      print "deleted" 
      connecting.close() 
      break 


def new_clients(): 
    from Class import Socket_control 
    while True: 
     connect = Socket_control.open_socket() 
     thread_listener = Thread(target=Socket_control.recv_msg, args=(connect,)) 
     thread_listener.start() 


def menu(): 
    choice = 1 
    while choice != 0: 
     choice = raw_input("What would you like to do?\n[0] exit.\n[1] print list of computers.") 
     if choice == "1": 
      print_my_socket_list() 
      if len(my_socket_list) == 0: 
       continue 
      computer = raw_input("Choose a computer :") 
      msg = raw_input("Send message: ") 
      my_socket_list[computer].send_message("MSG:" + msg) 


def main(): 

    new_client_thread = Thread(target=new_clients) 
    new_client_thread.start() 
    menu() 

my_socket_list = {} 
my_socket_list["0"] = None 

if __name__ == "__main__": 
    main()

这是我的插座控制

import socket 
import datetime 
import server 
from mutex import mutex 
def open_socket(): 
    global connect 
    try: 
     sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM) 
     sock.bind(("0.0.0.0", 8200)) 
    except socket.error: 
     print "Socket Error" 

    sock.listen(1) 
    connect, addr = sock.accept() 
    return connect 


def recv_msg(connecting): 
    while True: 
     try: 
      data = connecting.recv(1024) 
      if len(data) == 0: 
       server.connection_failed(connect) 
       return None 
      print "recv " + data 
      data = data.decode('utf-8') 
      if data[:4] == "MYID": 
       handle_client_id(data[4:]) 
     except: 
      server.connection_failed(connecting) 
      print "Recv Error" 
      return None 




def handle_client_id(data): 
    my_id = data 
    if my_id == "0": 
     my_id = get_last_number()+1 
     send_message("MYID" + str(my_id)) 
    file_writing_new_client(my_id) 
    server.add_client_to_dict(my_id,connect) 


def send_message(data): 
    try: 
     print "sent " + data 
     connect.send(data) 
    except: 
     print "Sent Error" 
     server.connection_failed(connect) 


def get_last_number(): 
    with open("jennifer_clients_list.txt") as my_file: 
     my_file.seek(0) 
     first_char = my_file.read(1) 
     my_file.seek(0) 
     if not first_char: 
      return 0 
     else: 
      lines = my_file.readlines() 
      print "text: " 
      print lines 
      data = lines[-1].split(" ")[0] 
      return int(data)

来源

2016-11-11 Rony Cohen

不能访问外部局部变量，需要使用'global'或'class'来访问任何地方。但是如果从外部访问变量，则会失去同步。最好的结果是存储结果比存储数据更好！ – dsgdfg

在CPython中，你应该没有问题，更新在一个线程的字典，而在另一个访问它（见Using a global dictionary with threads in Python）。

问题的近因是你试图用错误的键引用字典。当您将该ID存储在字典中时，首先将其转换为int，但是当您从主线程访问该ID时，您正在使用用户提供的str输入;即int('88')是而不是与'88'相同的密钥。

如果更改行：

my_socket_list[computer].send_message("MSG:" + msg)

到：

my_socket_list[int(computer)].send_message("MSG:" + msg)

你会进步到一个错误（这就是socket对象不具有send_message方法，但如果你改变它到.send，它会工作）。

要回到最初的主题，将代码留给实施细节（如CPython的全局解释器锁定）通常是一个坏主意，所以我建议同步访问您的字典Lock。

来源

2016-11-11 16:17:21

字典更新内部功能，但不在外面

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