Use this to select the right forum: https://groups.google.com/forum/?hl=en#!topic/android-discuss/BfKnSUV-7r0
Best regards, Paul. On Thu, Jul 21, 2011 at 3:54 PM, viny <[email protected]> wrote: > I have one application, to which many other services (from other > applications say 3rd party apps) can be subscribed. And I need to get > updated data into my application from the other applications, and so I > require have bi-directional communication from my application to other > services. The kind of data that I require from all the services is > same. > > Right now using Messenger and Bound Services serves my purpose for bi- > directional communication between my application and a service from > ONE other application. But now I am stuck how can I extend it to > support MULTIPLE services. > I know I can get the "ComponentName" in the callback function > "onServiceConnected(ComponentName, IBinder)" and so I can get the > corresponding IBinder interface from the corresponding service (using > just one ServiceConnection for all). But when we unbind any service, > then we unbind it with ServiceConnection as a parameter, so would it > unbind all of my subscribed services which are attached via that > ServiceConnection? Do I really need to have multiple > ServiceConnections for every service? I am getting really confused > with this. Could someone throw some light into possible solution? > > -- > You received this message because you are subscribed to the Google Groups > "Android Discuss" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/android-discuss?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Android Discuss" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/android-discuss?hl=en.
