问题描述
Main与单独模块定义的GUI进行交互。 Main中使用了两种替代方法来定义QShortcut的插槽。 lambda方法有效(但看起来很麻烦)。仅当先执行lambda方法时,direct方法才有效。如果使用lambda方法的行被注释掉,则直接方法将失败(无错误消息)。为什么仅直接方法会失败?
''' #p.py
import sys
from PyQt5 import QtWidgets as qtw
from Q import MainWindow
class MyActions():
def __init__(self,my_window):
self.my_window = my_window
self.my_window.run.activated.connect(lambda: self.rmssg1()) # had to use this as a work-around
self.my_window.run.activated.connect(self.rmssg2) # expected this to work
def rmssg1(self):
self.my_window.my_label1.setText('Ctrl+R pressed -> mssg 1')
def rmssg2(self):
self.my_window.my_label2.setText('Ctrl+R pressed -> mssg 2')
if __name__ == '__main__':
app = qtw.QApplication(sys.argv)
mw = MainWindow()
MyActions(mw)
mw.show()
sys.exit(app.exec())
''' 这是单独模块中的GUI
''' #q.py
import sys
from PyQt5 import QtWidgets as qtw
from PyQt5.QtGui import QKeySequence
class MainWindow(qtw.QMainWindow):
def __init__(self,parent=None):
super().__init__()
self.my_label1 = qtw.QLabel("Enter Ctrl+R",self)
self.my_label1.setGeometry(20,20,200,30)
self.my_label2 = qtw.QLabel("Enter Ctrl+R",self)
self.my_label2.setGeometry(20,50,30)
#define shortcut
self.run = qtw.QShortcut(QKeySequence('Ctrl+R'),self)
'''
解决方法
问题很简单:当您MyActions(mw)
不为该对象分配变量,从而导致内存丢失时,解决方法是:
if __name__ == '__main__':
app = qtw.QApplication(sys.argv)
mw = MainWindow()
foo = MyActions(mw)
mw.show()
sys.exit(app.exec())