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我遇到了套接字/ tkinter登录应用程序的问题。服务器在Raspberry Pi上,但即使服务器没有启动,我也没有理由认为这应该在启动时挂起。我已经导入了一个我将要包含的模块。使用套接字的Tkinter应用程序开始挂起
为什么它挂起?
这里是我的客户端代码 - 那挂着一个: -
import socket, pickle
import Tkinter as tk
import loginutility
class Server(object):
def __init__(self):
self.s = socket.socket()
self.p = 10000
self.ip = "192.168.1.120"
def addUser(self, userinfo):
puserinfo = pickle.dumps(userinfo)
self.s.connect((self.ip, self.p))
self.s.sendall("check")
self.sendall(puserinfo)
if self.s.recv(1024) == False:
self.s.sendall("add")
self.send(puserinfo)
return True
else:
return False
def userDump(self):
self.s.sendall("userdump")
return pickle.loads(self.s.recv(1024))
class Main(tk.Tk):
def __init__(self):
tk.Tk.__init__(self)
class LoginFrame(tk.Frame):
def __init__(self, parent):
tk.Frame.__init__(self, parent)
self.pack()
self.l = loginutility.LoginBox(self)
login = tk.Button(self, text="Login", command=self.login)
login.pack()
def login(self):
u, p = self.l.values()
users = Server.userDump()
if u in users and users[u] == p:
tk.Label(self, text="Success").pack()
main = Main()
mf = LoginFrame(main)
main.mainloop()
服务器代码进行测试时是不是:
import loginutility as lu
import socket
import pickle
s = socket.socket()
s.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
address = ''
port = 10000
s.bind((address, port))
s.listen(5)
users = lu.UserList("accounts")
while True:
c, clientaddress = s.accept()
c.send("You're Connected")
d = c.recv(1024)
if d == "add":
userdata = c.recv(1024)
username, password = pickle.loads(userdata)
users.adduser(username, password)
print "User {} added with password of {}.".format(username, password)
c.send("Success")
elif d == "check":
username, password = pickle.loads(c.recv(1024))
if users.checkuser(username) == False:
c.sendall(False)
else:
c.sendall(True)
elif d == "userdump":
c.send(pickle.dumps(users.dumpuser())
c.close()
必要loginutility代码:
class LoginBox:
def __init__(self, parent):
self.l1=Label(parent, text="Username:").grid()
self.ubox=Entry(parent)
self.ubox.grid()
self.l2=Label(parent, text="Password:").grid()
self.pbox=Entry(parent, show="*")
self.pbox.grid()
def values(self):
return (self.ubox.get(), self.pbox.get())
class UserList:
def __init__(self, file=None):
self.users = {}
self.file = file
if file != None:
with open(file, "rb") as f:
self.users = pickle.load(f)
def adduser(self, user, pswrd):
self.users[user] = pswrd
if file != None:
with open(self.file, "wb") as f:
pickle.dump(self.users, f)
def dumpuser(self):
return self.users
任何帮助非常感谢!
如果服务器代码在测试此代码时未启用,为什么要包含它?您应该将此代码降低到仍能再现问题的绝对最少线条。在这种情况下,你可能会删除2/3的代码。 –
您的服务器在测试时运行。当你调用self.s.connect()时,你的应用程序可能会永远等待你的服务器响应。如果您的服务器没有运行,客户端将冻结在那里。你可以通过添加一个[timeout](https://docs.python.org/2/library/socket.html#socket.socket.settimeout)到你的套接字来解决这个问题,你也可以通过做网络东西来避免锁定你的应用程序在另一个线程中。但要小心,因为线程会让你的应用程序**调试更复杂。 –
@TomásGonzalezDowling不,不是的,我已经接受⬇️下面的答案,但谢谢你的建议! –