【发布时间】:2019-01-18 19:59:45
【问题描述】:
不久前,我一直在使用(在工作中的给定框架中)Qt4 中的按名称自动连接插槽功能,并带有一个装饰器。像这样的:
self.button1 = QtWidgets.QPushButton("Click me!")
...
@QtCore.Slot()
def on_button1_clicked(self):
# whatever the method does...
而不是这个:
self.button1 = QtWidgets.QPushButton("Click me!")
self.button1.clicked.connect(self.handle_button)
...
def self.handle_button(self):
# whatever the method does...
现在我无法使其与 PySide2 (Qt-5.12) 一起使用。我在这里缺少什么使它起作用?
import sys
import random
from PySide2 import QtCore, QtWidgets, QtGui
class MyWidget(QtWidgets.QWidget):
def __init__(self):
QtWidgets.QWidget.__init__(self)
self.hello = ["Hallo Welt", "Hola Mundo"]
self.button1 = QtWidgets.QPushButton("Click me!")
self.text = QtWidgets.QLabel("Hello World")
self.text.setAlignment(QtCore.Qt.AlignCenter)
self.layout = QtWidgets.QVBoxLayout()
self.layout.addWidget(self.text)
self.layout.addWidget(self.button1)
self.setLayout(self.layout)
QtCore.QMetaObject.connectSlotsByName(self)
#self.button1.clicked.connect(self.on_button1_clicked)
@QtCore.Slot()
def on_button1_clicked(self):
self.text.setText(random.choice(self.hello))
if __name__ == "__main__":
app = QtWidgets.QApplication(sys.argv)
widget = MyWidget()
widget.show()
sys.exit(app.exec_())
【问题讨论】:
-
QtCore.QMetaObject.connectSlotsByName(self). -
您能说得更准确些吗?我在示例中做错了什么?
标签: python signals-slots pyside2