我如何将会话变量返回到PyQt5的主代码?

Question

我不知道如何在问题中表达我的问题，因为我是一个非常PyQt5的学习者，我不知道词汇，但是来吧。

我需要返回session_user变量(在login()函数中)到我的主代码。这个变量需要在登录完成后传递到主屏幕，但我不知道PyQt5中有一个函数返回一个不是int类型的变量:accept ()， reject ()， done()。

是否有可能将我的session_user变量返回到我的主代码中，使用一些PyQt5函数，就像我提到的那样?

我代码:

from PyQt5 import uic, QtWidgets
from PyQt5.QtWidgets import QDialog
from views.ui.login.telaLogin import Ui_Dialog
import sys
from control.exception_login import verification_login_user
class ViewLogin(QDialog):
def __init__(self):
super().__init__()
self.viewlogin = Ui_Dialog()
self.viewlogin.setupUi(self)
self.viewlogin.button_login.clicked.connect(self.login)
def login(self):
self.login = self.viewlogin.login_login.text() 
self.password = self.viewlogin.login_password.text()
erro =  verification_login_user(self.login, self.password)

if (erro == False):
self.close()
session_user = (self.login, self.senha) 
#self.done(r) # I understand its functionality # PROBLEM
#return (session_user) # PROBLEM
self.accept() # The problem would end here if I didn't need the variable sessao_user

elif(erro == True):
self.viewlogin.login_login.setText('')
self.viewlogin.login_password.setText('')

app = QtWidgets.QApplication([])
w = ViewLogin()
result = w.exec_()
print(result)  # desired result = (' login', 'password') 
# real result = 1 
sys.exit(app.exec_())

Answer 1

您可以通过threading模块实现的方法之一。

1. 创建类Controller和Daemon到True(这确保线程将在应用程序关闭后被杀死)。

class Controller(threading.Thread):
def __init__(self):
super(Controller, self).__init__(daemon = True)
self.my_variable = 0
self.variables_from_gui = None

2. 覆盖run方法:

def run(self):
print("My main loop is here!")
while True: 
print("Button presses: {}, Variables from gui: {}".format(self.button_presses, self.variables_from_gui))
time.sleep(1) #This wont freeze the gui

3. 添加button登录点击注册功能:

def login(self, variables_from_gui = None):
#Do you login stuff
self.variables_from_gui = variables_from_gui
self.button_presses += 1

4. 在App类中(在您的情况下，这将是主要的GUI类)，从login函数启动一个新的thread并将变量传递给它。

def login(self):
#Stuff to do in your GUI
variables_to_send = 0, "Text", {"key" : 10}
threading.Thread(target = lambda: self.controller.login(variables_from_gui = variables_to_send)).start()

5. 同样在你的GUI类中创建一个Controller对象并启动它:

self.controller = Controller()
self.controller.start() #Start the run method in the Controller()

---

有很多事情可以做得更好，使它更线程安全。但那不在这个问题的范围内。

整个代码:

---

import sys
from PyQt5.QtWidgets import QApplication, QWidget, QPushButton
from PyQt5.QtGui import QIcon
from PyQt5.QtCore import pyqtSlot
import threading
import time
class App(QWidget):
def __init__(self):
super().__init__()
self.left = 10
self.top = 10
self.width = 320
self.height = 200
self.initUI()

def initUI(self):
self.controller = Controller()
self.controller.start() #Start the run method in the Controller()
self.setGeometry(self.left, self.top, self.width, self.height)

button = QPushButton("My login", self)
button.move(100,70)
button.clicked.connect(self.login)

self.show()
def login(self):
#Stuff to do in your GUI
variables_to_send = 0, "Text", {"key" : 10}
threading.Thread(target = lambda: self.controller.login(variables_from_gui = variables_to_send)).start()
class Controller(threading.Thread):
def __init__(self):
super(Controller, self).__init__(daemon = True)
self.button_presses = 0
self.variables_from_gui = None
def run(self):
print("My main loop is here!")
while True: 
print("Button presses: {}, Variables from gui: {}".format(self.button_presses, self.variables_from_gui))
time.sleep(1) #This wont freeze the gui
def login(self, variables_from_gui = None):
#Do you login stuff
self.variables_from_gui = variables_from_gui
self.button_presses += 1

if __name__ == '__main__':
app = QApplication(sys.argv)
ex = App()
sys.exit(app.exec_())