wxPython: Sending a signal to several widgets

Question

I am not even sure how to ask this question. I want something that is like the wxPython event system, but a bit different. I'll try to explain.

When there is a certain change in my program (a "tree change", never mind what that is,) I want to send a signal to all the widgets in my program, notifying them that a "tree change" has occurred, and they should change their display in response.

How do I do this? It sounds a little bit like wxPython events, but not really, since events don't spread to all widgets, as far as I know.

What would be a good way to do this?

Anurag Uniyal · Accepted Answer · 2013-08-13 03:50:36Z

6

You can write your own publish-subscribe mechanism which can be as simple as this:

def register(self, callback):
    self.callbacks.append(callback)

def emit(self, eventName):
    for callback in self.callbacks:
         callback(eventName)

Anybody interested in listening to event registers a function with central registry and then you can emit a event to interested parties, you can improve it further by having to register for a specific event, having multiple registrars, unregister, error checking etc

Alternatively you can use wxPython's wx.lib.pubsub module or other python libraries like PyPubsub , PyDispatcher

edited Aug 13, 2013 at 3:50

answered Mar 31, 2010 at 4:11

Anurag Uniyal

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3 Comments

Steven Sproat Over a year ago

pubsub is the way to go imo :)

Oliver Over a year ago

@StevenSproat you should say why

Oliver Over a year ago

Note: PyPubSub listed in the answer is a standalone project but it is also distributed with wxPython as wx.lib.pubsub (which is none other than PyPubSub/src/pubsub from source distribution). PyPubSub has the advantage (over the register/emit example) of supporting hierarchies of messages, documentable message types, exception handling for listeners, and notification of pub-sub activity, features that you will find really important in larger applications.

Moayyad Yaghi · Accepted Answer · 2010-04-01 12:14:30Z

0

check out the observer design pattern. you need to implement the widgets as an observers . and the signal sender as the subject. so whenever it the subject sends a signal, all the observers will be notified.

check this out for more info about observers

answered Apr 1, 2010 at 12:14

Moayyad Yaghi

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