Python PyQt窗口运行x秒

Python PyQt窗口运行x秒,python,python-2.7,pyqt4,Python,Python 2.7,Pyqt4,是否可以使PyQt窗口保持打开x秒,然后通过 if __name__ == "__main__": import sys app = QtGui.QApplication(sys.argv) MainWindow = QtGui.QMainWindow() ui = Ui_MainWindow() ui.setupUi(MainWindow) MainWindow.show() sys.exit(app.exec_()) #enough c

是否可以使PyQt窗口保持打开x秒,然后通过

if __name__ == "__main__":
    import sys
    app = QtGui.QApplication(sys.argv)
    MainWindow = QtGui.QMainWindow()
    ui = Ui_MainWindow()
    ui.setupUi(MainWindow)
    MainWindow.show()
    sys.exit(app.exec_())
#enough code?

#pass to rest of the code?

如果我理解您的问题,解决方法是在传递关闭窗口的命令之前利用
time.sleep

import time
time.sleep(2) # delays for 2 seconds

您可以使用计时器关闭窗口

if __name__ == "__main__":
    import sys
    app = QtGui.QApplication(sys.argv)
    MainWindow = QtGui.QMainWindow()
    ui = Ui_MainWindow()
    ui.setupUi(MainWindow)
    MainWindow.show()
    QtCore.QTimer.singleShot(2000, MainWindow.close)
    sys.exit(app.exec_())

另一种方法是使用QDialog来显示ui,调用对话框的
exec\uu
,而不是QApplication

,即睡眠不允许Qt事件循环继续执行以保持窗口响应