simple IPython example raises exception on sys.exi

2019-01-09 07:27发布

问题:

I'm doing some very simple PySide (and PyQt) tutorials in IPython. One tutorial just creates a window with some sliders to demonstrate slots and signals.

When I close the window of the running demo application, I see this error:

An exception has occurred, use %tb to see the full traceback.
SystemExit: 0
To exit: use 'exit', 'quit', or Ctrl-D.

So I run %tb and get this:

SystemExit                                Traceback (most recent call last)
/Workspaces/scratch/<ipython-input-1-88966dcfb499> in <module>()
     33 
     34 if __name__ == "__main__":
---> 35     main()

/Workspaces/scratch/<ipython-input-1-88966dcfb499> in main()
     29         w.show()
     30         app.exec_()
---> 31         sys.exit(0)
     32 
     33 

SystemExit: 0

If I try to execute my code again, I get this:

RuntimeError: A QApplication instance already exists.

In case it helps, here my code:

from PySide.QtCore import *
from PySide.QtGui import *
import sys

class MyWindow(QWidget):
    def __init__(self):
        QWidget.__init__(self, None)

        vbox = QVBoxLayout(self)

        self.slider1 = QSlider(Qt.Horizontal)
        self.slider1.setRange(0, 99)
        self.slider1.setValue(0)
        vbox.addWidget(self.slider1)

        self.slider2 = QSlider(Qt.Horizontal)
        self.slider2.setRange(0, 99)
        self.slider2.setValue(99)
        vbox.addWidget(self.slider2)

        self.slider1.valueChanged.connect(self.slider2Changed)

    def slider2Changed(self, position):
        self.slider2.setValue(self.slider2.maximum() - position)

def main():
        app = QApplication(sys.argv)
        w = MyWindow()
        w.show()
        app.exec_()
        sys.exit(0)

if __name__ == "__main__":
    main()

I do not have errors when running the code using python:

python myexample.py

This error only happens when I run the code in an IPython (including a notebook or the qtconsole or the regular ipython terminal).

UPDATE: My main problem is that I cannot run the application again quickly and easily. If I try to run my code again, I get this:

RuntimeError: A QApplication instance already exists.

That kills the fast, interactive nature of IPython :(

回答1:

What you need to do is to cause the QApplication to be deleted later as in:

app = QApplication(sys.argv)
app.aboutToQuit.connect(app.deleteLater)

Using this code you can rerun the application as many times as you want in IPython, or anywhere else and every time you close your qt application, the object will be deleted in python.



回答2:

This answer is thanks to Matthias BUSSONNIER from the ipython-users mailing list.

When I close the window of the running demo application, I see this error: An exception has occurred, use %tb to see the full traceback. SystemExit: 0

Just don't use sys.exit(0) as you are not exiting python, but still running IPython.

Add it back if you wish to run your app from a (real) command line and have a return status.

If I try to execute my code again, I get this:
RuntimeError: A QApplication instance already exists.

This is a PySide Bug that they "won't fix" as they don't consider it a bug.

See https://github.com/ipython/ipython/issues/1124)
and http://bugs.pyside.org/show_bug.cgi?id=855

QApplication can only have one instance and quitting an app is apparently not considered a reason sufficient enough do delete the object...

You can use this code from above issues :

app=QtGui.QApplication.instance() # checks if QApplication already exists 
if not app: # create QApplication if it doesnt exist 
     app = QtGui.QApplication(sys.argv)

This was a sufficient solution for my present needs.



回答3:

sys.exit just raises SystemExit to terminate the interperter.

ipython catches SysExit when it executes a script in interactive mode, so this isn't acutally an error but a feature of ipython do avoid the interactive interpreter from being shutdown when a script is executed, as that's not what you usually want in an interactive session.