我一直在写一个在服务器上运行远程脚本的程序。所以,我需要用一个条来显示进度,但不知怎么的,当我运行代码时,GUI开始冻结。我使用了QThread和SIGNAL,但不幸的是无法成功。
下面是我的代码
class dumpThread(QThread):
def __init__(self):
QThread.__init__(self)
def __del__(self):
self.wait()
def sendEstablismentCommands(self, connection):
# Commands are sending sequently with proper delay-timers #
connection.sendShell("telnet localhost 21000")
time.sleep(0.5)
connection.sendShell("admin")
time.sleep(0.5)
connection.sendShell("admin")
time.sleep(0.5)
connection.sendShell("cd imdb")
time.sleep(0.5)
connection.sendShell("dump subscriber")
command = input('$ ')
def run(self):
# your logic here
# self.emit(QtCore.SIGNAL('THREAD_VALUE'), maxVal)
self.sendEstablismentCommands(connection)
class progressThread(QThread):
def __init__(self):
QThread.__init__(self)
def __del__(self):
self.wait()
def run(self):
# your logic here
while 1:
maxVal = 100
self.emit(SIGNAL('PROGRESS'), maxVal)
class Main(QtGui.QMainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self)
self.ui = Ui_MainWindow()
self.ui.setupUi(self)
self.ui.connectButton.clicked.connect(self.connectToSESM)
def connectToSESM(self):
## Function called when pressing connect button, input are being taken from edit boxes. ##
## dumpThread() method has been designed for working thread seperate from GUI. ##
# Connection data are taken from "Edit Boxes"
# username has been set as hardcoded
### Values Should Be Defined As Global ###
username = "ntappadm"
password = self.ui.passwordEdit.text()
ipAddress = self.ui.ipEdit.text()
# Connection has been established through paramiko shell library
global connection
connection = pr.ssh(ipAddress, username, password)
connection.openShell()
pyqtRemoveInputHook() # For remove unnecessary items from console
global get_thread
get_thread = dumpThread() # Run thread - Dump Subscriber
self.progress_thread = progressThread()
self.progress_thread.start()
self.connect(self.progress_thread, SIGNAL('PROGRESS'), self.updateProgressBar)
get_thread.start()
def updateProgressBar(self, maxVal):
for i in range(maxVal):
self.ui.progressBar.setValue(self.ui.progressBar.value() + 1)
time.sleep(1)
maxVal = maxVal - 1
if maxVal == 0:
self.ui.progressBar.setValue(100)
def parseSubscriberList(self):
parsing = reParser()
def done(self):
QtGui.QMessageBox.information(self, "Done!", "Done fetching posts!")
if __name__ == "__main__":
app = QtGui.QApplication(sys.argv)
main = Main()
main.show()
sys.exit(app.exec_())
我希望看到updateProgressBar方法已使用信号调用,所以进程将通过单独的线程。我找不到我失踪的地方。
谢谢你的帮助
有两个问题。我注意到的一件事是,如果Python线程不用于IO操作(比如从串行端口读取),那么它们是贪婪的。如果告诉线程运行计算或与IO无关的操作,则线程将占用所有处理,并且不希望让主线程/事件循环运行。第二个问题是信号很慢。。。非常慢。我注意到,如果你从一个线程发出一个信号并且做得非常快,它会大大减慢程序的速度。
所以问题的核心是,线程占用了所有的时间,而你发出的信号非常快,这将导致减速。
为了清晰和易于使用,我会使用新的风格的信号和插槽。 http://pyqt.sourceforge.net/Docs/PyQt4/new_style_signals_slots.html
连接到新样式信号
编辑 抱歉,还有一个问题。当一个信号调用一个函数时,你不能永远留在那个函数中。该函数没有在单独的线程中运行,而是在主事件循环上运行,主事件循环等待运行,直到您退出该函数。
更新进度休眠1秒并保持循环。绞刑是因为在这个函数中停留。
最好像这样写进度条
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