問題描述

我在一個項目上工作:程序下載但我在使用 while 循環時遇到問題，用于檢查與 Internet 的連接，如果 true 不 setText('') to lable 和 if Flase setText('anyText') to lable

I work on one Project : Program download but I have a problem with while loop for check the connection with the internet and if true doesn't setText('') to lable and if Flase setText('anyText') to lable

連接檢查方法

    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()

我已經厭倦了 QThread .請問我該怎么做:)?如果連接丟失=False setText('check internet') 以及連接變為 true 時，應用程序正在運行時 setText('')

I have tired with QThread . Please How I can do it :) ? And when app is running if connection is lost=False setText('check internet') and when the connection become true setText('')

構造者

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

主代碼

def main():
    app = QApplication(sys.argv)
    window = mainApp()
    window.checkInternetConnection()
    window.show()
    app.exec()

if __name__=='__main__':
    main()

推薦答案

QThread不要太復雜，使用線程庫:

Do not get complicated with QThread, use the threading library:

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，由于您使用的是線程，因此不應從另一個線程更新 GUI，為此您可以使用 QMetaObject::invokeMethod:

On the other hand, since you are using a thread, you should not update the GUI from another thread, for this you can use 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()

這篇關于如何使用 PyQt5 運行 while 循環的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！