python - 信号和函数之间的 PyQt 程序连接不起作用

Question

在这个非常简单的基于 PyQt 的 Python 程序中 start按钮不起作用，开始按钮和_calculateResult之间似乎没有联系方法。我想在 _connectSignals()方法是错误的，但我找不到它。你有什么想法吗？谢谢。

import sys

from PyQt5.QtCore import Qt
from PyQt5.QtWidgets import QApplication, QMainWindow, QWidget, QLineEdit, QPushButton, QHBoxLayout

class TimerUi(QMainWindow):
    def __init__(self):
        super().__init__()
        self.setWindowTitle('Basic Timer')
        self.setFixedSize(235, 235)
        self.generalLayout = QHBoxLayout()
        self._centralWidget = QWidget(self)
        self.setCentralWidget(self._centralWidget)
        self._centralWidget.setLayout(self.generalLayout)
        self._createDisplayAndButtons()

    def _createDisplayAndButtons(self):
        self.display = QLineEdit()
        self.display.setFixedHeight(35)
        self.generalLayout.addWidget(self.display)

        self.buttons = {}
        self.buttons['start'] = QPushButton('Start')
        self.buttons['start'].setFixedSize(40, 40)
        self.generalLayout.addWidget(self.buttons['start'])

    def setDisplayText(self, text):
        self.display.setText(text)
        self.display.setFocus()

class PyCalcCtrl:
    def __init__(self, view):
        self._view = view
        self._connectSignals()

    def _calculateResult(self):
        self._view.setDisplayText('Time is 17:13')

    def _connectSignals(self):
        self._view.buttons['start'].clicked.connect(self._calculateResult)

def main():
    basic_timer = QApplication(sys.argv)
    view = TimerUi()
    view.show()
    PyCalcCtrl(view=view)
    sys.exit(basic_timer.exec_())


if __name__ == '__main__':
    main()

Answer 1

通过创建 PyCalcCtrl 对象而不将其分配给变量，GC 将其删除。解决方案是将该对象分配给一个变量:

def main():
    basic_timer = QApplication(sys.argv)
    view = TimerUi()
    view.show()
    ctrl = PyCalcCtrl(view=view)
    sys.exit(basic_timer.exec_())