为了使我的同事更加安全，并向他们展示即使使用最简单的编码来破解密码也是多么容易，我编写了一个简单的程序，仅用蛮力检查每种可能的组合。 为了给他们一个玩耍的小工具，我想将其包装在一个小的pyqt应用程序中。 在Spyder和控制台上，即使使用更长的密码，破解程序也可以正常运行。但是，当我尝试测试更长的密码时，使用PyQT5会死机。例如。在控制台上，破解abcd34大约需要1:30分钟，即使10分钟后也没有使用PyQt。 我关注了这篇文章（Pyqt5 qthread + signal not working + gui freeze）并使用了线程，但是它一直处于冻结状态。有什么想法我做错了吗？
谢谢
安雅
这是我的代码：
import itertools
import string
import datetime as dt
from PyQt5.QtCore import QObject, QThread, pyqtSignal, pyqtSlot
from PyQt5.QtWidgets import QApplication, QPushButton, QTextEdit, QVBoxLayout, QWidget, QLineEdit
def trap_exc_during_debug(*args):
# when app raises uncaught exception, print info
print(args)
# install exception hook: without this, uncaught exception would cause application to exit
sys.excepthook = trap_exc_during_debug
class Worker(QObject):
"""
Must derive from QObject in order to emit signals, connect slots to other signals, and operate in a QThread.
"""
sig_done = pyqtSignal(int) # worker id: emitted at end of work()
sig_msg = pyqtSignal(str) # message to be shown to user
def __init__(self, id: int):
super().__init__()
self.__id = id
self.__abort = False
@pyqtSlot()
def work(self, value):
"""
Pretend this worker method does work that takes a long time. During this time, the thread's
event loop is blocked, except if the application's processEvents() is called: this gives every
thread (incl. main) a chance to process events, which in this sample means processing signals
received from GUI (such as abort).
"""
thread_name = QThread.currentThread().objectName()
thread_id = int(QThread.currentThreadId()) # cast to int() is necessary
self.sig_msg.emit('Starting brute force password cracking.\n\n')
time1 = dt.datetime.now()
special = "@%-*,'&`._!#$§äöü?=+:;/\[]{}()"
chars = string.ascii_letters + string.digits + special
attempts = 0
for password_length in range(1, 10):
for guess in itertools.product(chars, repeat=password_length):
attempts += 1
guess = ''.join(guess)
if guess == value: #self.freetextle.text():
time2 = (dt.datetime.now() - time1)
self.sig_done.emit(self.__id)
return self.sig_msg.emit('Password is {} and it was found in {} guesses, which took me {} seconds'.format(guess, attempts, time2))
# check if we need to abort the loop; need to process events to receive signals;
app.processEvents() # this could cause change to self.__abort
if self.__abort:
# note that "step" value will not necessarily be same for every thread
self.sig_msg.emit('aborting')
self.sig_done.emit(self.__id)
def abort(self):
self.sig_msg.emit('notified to abort')
self.__abort = True
class MyWidget(QWidget):
# sig_start = pyqtSignal() # needed only due to PyCharm debugger bug (!)
sig_abort_workers = pyqtSignal()
def __init__(self):
super().__init__()
self.setWindowTitle("Simple password cracker")
form_layout = QVBoxLayout()
self.setLayout(form_layout)
self.resize(400, 400)
self.button_start_threads = QPushButton()
self.button_start_threads.clicked.connect(self.start_threads)
self.button_start_threads.setText("Start")
form_layout.addWidget(self.button_start_threads)
# create the input field
self.freetextle = QLineEdit()
self.freetextle.setObjectName("password")
self.freetextle.setText("Enter a password to check")
form_layout.addWidget(self.freetextle)
self.log = QTextEdit()
form_layout.addWidget(self.log)
QThread.currentThread().setObjectName('main') # threads can be named, useful for log output
self.__workers_done = None
self.__threads = None
def start_threads(self):
#self.log.append('starting threads')
self.button_start_threads.setDisabled(True)
self.__workers_done = 0
self.__threads = []
worker = Worker(1)
thread = QThread()
thread.setObjectName('thread_1')# + str(idx))
self.__threads.append((thread, worker)) # need to store worker too otherwise will be gc'd
worker.moveToThread(thread)
# get progress messages from worker:
worker.sig_step.connect(self.on_worker_step)
worker.sig_done.connect(self.on_worker_done)
worker.sig_msg.connect(self.log.append)
# control worker:
self.sig_abort_workers.connect(worker.abort)
# get read to start worker:
# self.sig_start.connect(worker.work) # needed due to PyCharm debugger bug (!); comment out next line
thread.started.connect(worker.work(self.freetextle.text()))
thread.start() # this will emit 'started' and start thread's event loop
# self.sig_start.emit() # needed due to PyCharm debugger bug (!)
@pyqtSlot(int, str)
def on_worker_step(self, worker_id: int, data: str):
self.log.append('Worker #{}: {}'.format(worker_id, data))
@pyqtSlot(int)
def on_worker_done(self, worker_id):
self.log.append('done')
self.__workers_done += 1
#if self.__workers_done == 1:
# self.log.append('No more workers active')
self.button_start_threads.setEnabled(True)
# self.__threads = None
@pyqtSlot()
def abort_workers(self):
self.sig_abort_workers.emit()
self.log.append('Asking to abort')
for thread, worker in self.__threads: # note nice unpacking by Python, avoids indexing
thread.quit() # this will quit **as soon as thread event loop unblocks**
thread.wait() # <- so you need to wait for it to *actually* quit
# even though threads have exited, there may still be messages on the main thread's
# queue (messages that threads emitted before the abort):
self.log.append('All threads exited')
if __name__ == "__main__":
app = QApplication([])
form = MyWidget()
form.show()
sys.exit(app.exec_())
答案 0 :(得分：0)
您的问题在thread.started.connect中。在那里，您可以通过传递参数来调用工作方法，而不仅仅是建立信号/插槽连接。
鉴于开始时没有参数，则无法通过此连接将参数传递给插槽。因此，您需要将文本作为参数传递给worker构造函数。并在Worker中将其用作成员。