我的套接字程序挂起在clientsocket,address)=服务器套接字。接受()也不会说我们有错什么的。你知道吗
我遵循https://docs.python.org/3/howto/sockets.html的指示
我已经想了一个小时了,但是没有结果。顺便说一句,我用的是python3。我做错什么了?编辑:我的遗嘱是完全搞砸了,因为我把它贴错了,但除此之外,我的代码是我在我的文件中。你知道吗
#import socket module
import socket
#creates an inet streaming socket.
serversocket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
print('socket created')
#binds socket to a public host, and a well known port
serversocket.bind(('127.0.0.1', 1024))
#print(socket.gethostname())# on desktop prints 'myname-PC')
#become a server socket
serversocket.listen(5) # listens for up to 5 requests
while True:
#accept connections from outside
#print('In while true loop') This works, but we never get to the next print statement. Why the hell is it catching at line 20?
(clientsocket, address) = serversocket.accept()
#clientsocket = serversocket.accept()
print('Ready to serve')
#now we do something with client socket...
try:
message = clientsocket.recv(1024)
filename = message.split()[1]
f = open(filename[1:])
outputdata = f.read()
#send an http header line
clientsocket.send('HTTP/1.1 200 OK\nContent-Type: text/html\n\n')
for i in range(0, len(outputdata)):
clientsocket.send(outputdata[i])
clientsocket.close()
except IOERROR:
clientsocket.send('HTTP/1.1 404 File not found!')
clientsocket.close()
如果您还没有编写客户端脚本/程序来连接到套接字并向其发送数据,那么它也会挂起服务器套接字。接受()因为没有什么可接受的。但假设你有。。。你知道吗
它挂起是因为由于True总是True,所以循环永远不会退出。在所提供的示例中,一旦连接被接受,它们就假装服务器是线程化的,其思想是创建一个单独的线程来开始读取和处理接收到的数据,从而允许套接字继续侦听更多的连接。你知道吗
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