如何在PyQT中的类之间连接pyqtSignal

2024-05-29 03:31:51 发布

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如何在两个不同的对象(类)之间正确地连接pyqtSignal?我是说最佳实践。

看看我为实现这个目标所做的:当Pot温度升高时,Thermometer类会得到通知:

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/更好的方法来做吗?

我还坚持(如果可能的话)使用“新”样式的PyQt信号。


Tags: selfobjinitdefclassparenttemperaturepot
1条回答
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1楼 · 发布于 2024-05-29 03:31:51
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()

这就是我根据文档所做的:
http://www.riverbankcomputing.com/static/Docs/PyQt4/html/new_style_signals_slots.html

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