用Python创建WebSocket客户端

Question

我正在尝试学习套接字编程以及WebSocket协议.我知道存在python web socket客户端,但我希望只为自己的学习构建一个玩具版本.为此,我创建了一个非常简单的Tornado websocket服务器,我正在运行localhost:8888.它只是在客户端连接时打印一条消息.

这是整个服务器 - 它的工作原理(我在浏览器中用一个小的javascript脚本测试过它)

import tornado.httpserver
import tornado.websocket
import tornado.ioloop
import tornado.web


class WSHandler(tornado.websocket.WebSocketHandler):
    def open(self):
        print('new connection')
        self.write_message("Hello World")

    def on_message(self, message):
        print('message received %s' % message)

    def on_close(self):
      print('connection closed')

application = tornado.web.Application([
    (r'/ws', WSHandler),
])


if __name__ == "__main__":
    http_server = tornado.httpserver.HTTPServer(application)
    http_server.listen(8888)
    tornado.ioloop.IOLoop.instance().start()

所以一旦我启动服务器,我尝试运行以下脚本

import socket

sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.connect((socket.gethostbyname('localhost'), 8888))

msg = '''GET /chat HTTP/1.1
Host: server.example.com
Upgrade: websocket
Connection: Upgrade
Sec-WebSocket-Key: dGhlIHNhbXBsZSBub25jZQ==
Origin: http://example.com
Sec-WebSocket-Protocol: chat, superchat
Sec-WebSocket-Version: 13'''.encode('ascii')
print(len(msg))

sent_count = sock.send(msg)
print('sent this many bytes:', sent_count)
recv_value = sock.recv(1)
print('recvieved:', recv_value)

我希望服务器将发回RFC中指定的响应头.相反,sock.recv挂了.这让我相信服务器没有确认websocket初始握手.此握手也是从RFC中撤出的.我知道websocket密钥应该是随机的一切,但我认为这不会导致服务器忽略握手(websocket密钥有效).我想,一旦我能够发起握手,我就能解决其余问题,所以我希望在websockets的工作方式或如何发送初始握手方面存在一些误解.

Answer 1

1)当您通过套接字发送消息时,您不知道它将被分成多少块.它可能都会立即发送; 或者可以发送前3个字母,然后发送剩余的信息; 或者消息可以分成10个部分.

2)给定1)服务器如何知道何时收到客户端发送的所有块？例如,假设服务器收到客户端消息的1个块.服务器如何知道这是整个消息还是还有9个以上的块？

3)我建议你读这个:

http://docs.python.org/2/howto/sockets.html

(加上评论中的链接)

4)现在,为什么不使用python来创建HTTP服务器？

python3:

import http.server
import socketserver

PORT = 8000
handler = http.server.SimpleHTTPRequestHandler

httpd = socketserver.TCPServer(("", PORT), handler)

print("serving at port", PORT)
httpd.serve_forever()

python2:

import SimpleHTTPServer
import SocketServer

PORT = 8000
handler = SimpleHTTPServer.SimpleHTTPRequestHandler

httpd = SocketServer.TCPServer(("", PORT), handler)

print "serving at port", PORT
httpd.serve_forever()

SimpleHTTPRequestHandler提供服务器程序目录及其下的文件,将请求URL与您创建的目录结构相匹配.如果请求'/',服务器将从服务器所在的目录中提供index.html文件.以下是python 3的客户端套接字示例(下面是python 2示例):

import socket   
import sys

try:
    s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
except socket.error:
    print('Failed to create socket')
    sys.exit()

print('Socket Created')

#To allow you to immediately reuse the same port after 
#killing your server:
s.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)

host = 'localhost';
port = 8000;

s.connect((host , port))

print('Socket Connected to ' + host + ' on port ', port)


#Send some data to server
message = "GET / HTTP/1.1\r\n\r\n"

try :
    #Send the whole string(sendall() handles the looping for you)
    s.sendall(message.encode('utf8') )
except socket.error:
    print('Send failed')
    sys.exit()

print('Message sent successfully')

#Now receive data
data = [] 

while True:
    chunk = s.recv(4096)  #blocks while waiting for data
    if chunk: data.append(chunk.decode("utf8"))
    #If the recv() returns a blank string, then the other side
    #closed the socket, and no more data will be sent:
    else: break  

print("".join(data))

--output:--
Socket Created
Socket Connected to localhost on port  8000
Message sent successfully
HTTP/1.0 200 OK
Server: SimpleHTTP/0.6 Python/3.2.3
Date: Sat, 08 Jun 2013 09:15:18 GMT
Content-type: text/html
Content-Length: 23
Last-Modified: Sat, 08 Jun 2013 08:29:01 GMT

<div>hello world</div>

在python 3中,你必须使用带有套接字的字节字符串,否则你会得到可怕的:

TypeError: 'str' does not support the buffer interface

这是在python 2.x中:

import socket   
import sys

try:
    s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
except socket.error:
    print 'Failed to create socket'
    sys.exit()

print('Socket Created')

#To allow you to immediately reuse the same port after 
#killing your server:
s.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)

host = 'localhost';
port = 8000;

s.connect((host , port))

print('Socket Connected to ' + host + ' on port ', port)

#Send some data to server
message = "GET / HTTP/1.1\r\n\r\n"

try :
    #Send the whole string(handles the looping for you)
    s.sendall(message)
except socket.error:
    print 'Send failed'
    sys.exit()

print 'Message sent successfully'

#Now receive data
data = [] 

while True:
    chunk = s.recv(4096)  #blocks while waiting for data
    if chunk: data.append(chunk)
    #If recv() returns a blank string, then the other side
    #closed the socket, and no more data will be sent:
    else: break  

print("".join(data))

--output:--
Message sent successfully
HTTP/1.0 200 OK
Server: SimpleHTTP/0.6 Python/2.7.3
Date: Sat, 08 Jun 2013 10:06:04 GMT
Content-type: text/html
Content-Length: 23
Last-Modified: Sat, 08 Jun 2013 08:29:01 GMT

<div>hello world</div>

请注意,GET请求的标头告诉服务器HTTP 1.1将是协议,即管理会话的规则.正如HTTP 1.1的RFC所描述的那样,请求中必须有两个'\ r \n'序列.所以服务器正在寻找第二个'\ r \n'序列.如果从请求中删除其中一个'\ r \n'序列,客户端将挂起recv(),因为服务器仍在等待更多数据,因为服务器还没有读取第二个'\ r \n'序列.

另请注意,您将以字节形式发送数据(在python 3中),因此不会有任何自动'\n'转换,并且服务器将期望序列'\ r \n'.