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[util.popup] Deduplicate code, "close" button only if leave=False

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Naya Verdier 2020-07-25 08:43:26 -07:00
parent d358790c6a
commit 4aff0499f0
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GPG key ID: 1A59389D46A94A4C
1 changed files with 10 additions and 14 deletions
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bumblebee_status/util/popup.py
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@ -16,23 +16,19 @@ class menu(object):
def __init__(self, parent=None, leave=True):
self.running = True
self.parent = None
if not parent:
self._root = tk.Tk()
self._root.withdraw()
self._menu = tk.Menu(self._root, tearoff=0)
self._menu.bind("<FocusOut>", self.__on_focus_out)
self.add_menuitem("close", self.__on_focus_out)
self.add_separator()
else:
self._root = parent.root()
self._root.withdraw()
self._menu = tk.Menu(self._root, tearoff=0)
self._menu.bind("<FocusOut>", self.__on_focus_out)
self.parent = parent
self.parent = parent
self._root = parent.root() if parent else tk.Tk()
self._root.withdraw()
self._menu = tk.Menu(self._root, tearoff=0)
self._menu.bind("<FocusOut>", self.__on_focus_out)
if leave:
self._menu.bind("<Leave>", self.__on_focus_out)
elif not parent:
self.add_menuitem("close", self.__on_focus_out)
self.add_separator()
self._menu.bind("<ButtonRelease-1>", self.release)