我在一个项目上工作:程序下载,但是用while循环检查与互联网的连接是否存在问题,如果为true,则setText('')不会变为lable,而Flase setText('anyText')变为able
检查连接的方法
def checkInternetConnection(self,host="8.8.8.8", port=53, timeout=3):
while self.conection==False:
try:
socket.setdefaulttimeout(timeout)
socket.socket(socket.AF_INET, socket.SOCK_STREAM).connect((host, port))
self.conection = True
return self.conection
except Exception as e:
print(e)
self.label_9.setText('Please Check Internect Connection')
self.conection = False
return self.conection
self.finished.emit()
我对QThread感到厌倦。请我怎么做:)?当应用程序运行时,如果连接丢失= False setText('check internet'),并且当连接变为true时,setText('')
构造器
From_Class,_=loadUiType(os.path.join(os.path.dirname(__file__),'designe.ui'))
class mainApp(QMainWindow,From_Class):
finished = pyqtSignal()
def __init__(self,parent=None):
super(mainApp, self).__init__(parent)
QMainWindow.__init__(self)
super().setupUi(self)
self.handleGui()
self.handleButton()
self.setWindowIcon(QIcon('mainIcon.png'))
self.menuBarW()
self.conection = False
MainCode
def main():
app = QApplication(sys.argv)
window = mainApp()
window.checkInternetConnection()
window.show()
app.exec()
if __name__=='__main__':
main()
答案 0 :(得分:0)
不要对QThread感到复杂,请使用线程库:
def main():
app = QtWidgets.QApplication(sys.argv)
window = mainApp()
threading.Thread(target=window.checkInternetConnection, daemon=True).start()
window.show()
app.exec()
另一方面,由于您正在使用线程,因此不应从其他线程更新GUI,为此您可以使用QMetaObject::invokeMethod
:
def checkInternetConnection(self,host="8.8.8.8", port=53, timeout=3):
while True:
try:
socket.setdefaulttimeout(timeout)
socket.socket(socket.AF_INET, socket.SOCK_STREAM).connect((host, port))
self.conection = True
except Exception as e:
self.conection = False
print(e)
msg = "" if self.conection else 'Please Check Internect Connection'
print("msg", msg)
QtCore.QMetaObject.invokeMethod(self.label_9, "setText",
QtCore.Qt.QueuedConnection,
QtCore.Q_ARG(str, msg))
self.finished.emit()