如何在 PyQT 中的类之间连接 pyqtSignal [英] How to connect pyqtSignal between classes in PyQT
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问题描述
如何正确连接两个不同的对象(类)之间的 pyqtSignal
?我的意思是最佳实践.
How to connect pyqtSignal
between two different objects (classes) PROPERLY? I mean best practice.
看看我为实现目标做了什么:Thermometer
类在 Pot
温度升高时收到通知:
Look what I have done to achieve the goal: The Thermometer
class is notified when Pot
increases its temperature:
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()
或者有什么更简单/更好的方法吗?
Or is there any easier / better way to do it?
我也坚持(如果可能的话)使用新"风格的 PyQt 信号.
I also insist (if possible) on using "new" style PyQt signals.
推荐答案
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()
根据文档,我会这样做:
http://www.riverbankcomputing.com/static/Docs/PyQt4/html/new_style_signals_slots.html
This is how I would've done it according to the docs:
http://www.riverbankcomputing.com/static/Docs/PyQt4/html/new_style_signals_slots.html
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