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I have created a PyGTK application that shows a Dialog when the user presses a button. The dialog is loaded in my __init__ method with: builder = gtk.Builder() builder.add_from_file("filename") builder.connect_signals(self) self.myDialog = builder.get_object("dialog_name") In the event handler, the dialog is shown with the command self.myDialog.run() , but this only works once, because after run() the dialog is automatically destroyed. If I click the button a second time, the application crashes. I read that there is a way to use show() instead of run() where the dialog is not destroyed, but I feel like this is not the right way for me because I would like the dialog to behave modally and to return control to the code only after the user has closed it. Is there a simple way to repeatedly show a dialog using the run() method using gtkbuilder? I tried reloading the whole dialog using the gtkbuilder, but that did not really seem to work, the dialog was missing all child elements (and I would prefer to have to use the builder only once, at the beginning of the program). [SOLUTION] (edited) As pointed out by the answer below, using hide() does the trick. I first thought you still needed to catch the "delete-event", but this in fact not necessary. A simple example that works is: import pygtk import gtk class DialogTest: def rundialog(self, widget, data=None): self.dia.show_all() result = self.dia.run() self.dia.hide() def destroy(self, widget, data=None): gtk.main_quit() def __init__(self): self.window = gtk.Window(gtk.WINDOW_TOPLEVEL) self.window.connect("destroy", self.destroy) self.dia = gtk.Dialog('TEST DIALOG', self.window, gtk.DIALOG_MODAL | gtk.DIALOG_DESTROY_WITH_PARENT) self

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