如何在PyQT中的类之间连接pyqtSignal

时间:2010-10-08 14:17:53

标签: python pyqt pyqt4

如何正确地在两个不同的对象(类)之间连接pyqtSignal?我的意思是最佳实践。

了解我为实现目标所做的工作:Thermometer类会在Pot温度升高时收到通知:

from PyQt4 import QtCore

class Pot(QtCore.QObject):
    temperatureRaisedSignal = QtCore.pyqtSignal()

    def __init__(self, parent=None):
        super(Pot, self).__init__(parent)
        self.temperature = 1
    def Boil(self):
        self.temperature += 1
        self.temperatureRaisedSignal.emit()
    def RegisterSignal(self, obj):
        self.temperatureRaisedSignal.connect(obj)

class Thermometer():
    def __init__(self, pot):
        self.pot = pot
        self.pot.RegisterSignal(self.temperatureWarning)
    def StartMeasure(self):
        self.pot.Boil()
    def temperatureWarning(self):
        print("Too high temperature!")

if __name__ == '__main__':
    pot = Pot()
    th = Thermometer(pot)
    th.StartMeasure()

或者有更简单/更好的方法吗?

我也坚持(如果可能的话)使用“新”式PyQt信号。

1 个答案:

答案 0 :(得分:21)

from PyQt4 import QtCore

class Pot(QtCore.QObject):

    temperatureRaisedSignal = QtCore.pyqtSignal()

    def __init__(self, parent=None):
        QtCore.QObject.__init__(self)
        self.temperature = 1

    def Boil(self):
        self.temperatureRaisedSignal.emit()
        self.temperature += 1

class Thermometer():
    def __init__(self, pot):
        self.pot = pot
        self.pot.temperatureRaisedSignal.connect(self.temperatureWarning)

    def StartMeasure(self):
        self.pot.Boil()

    def temperatureWarning(self):
        print("Too high temperature!")

if __name__ == '__main__':
    pot = Pot()
    th = Thermometer(pot)
    th.StartMeasure()

这就是我根据文档完成它的方式:
http://www.riverbankcomputing.com/static/Docs/PyQt4/html/new_style_signals_slots.html