I just guess, Whole the services are inside of the same process,(as you said), no service will be killed by android automatically.
Am I right ? :-) 2009/12/16 Carter <[email protected]> > *bump* > > Again, the question is: if there are multiple services within a > process and at least one service is a "foreground" service, will that > keep the non-foreground services alive? > > > Thanks, > Carter > > The documentation > > On Nov 23, 10:40 pm, Carter <[email protected]> wrote: > > This is probably a question for Dianne: > > > > The Android 2.0 SDK introduces the startForeground() API which makes > > it much less likely that a background service (and therefore a given > > process) will be killed. > > > > Suppose there are two services within a process, and one of those > > services has called startForeground() while the other has not. Will > > the one foreground service "protect" the other non-foreground service > > from being killed? > > > > I'm specifically looking at a case where there could be multiple > > services within a process, but coming from different APKs. So these > > would be APKs signed with the same signature, and given the same > > sharedUserID and process in the Android Manifest. I'd like to avoid > > having to post a notification for each individual service, as that > > would likely annoy users. > > > > Thanks! > > Carter > > -- > You received this message because you are subscribed to the Google > Groups "Android Developers" group. > To post to this group, send email to [email protected] > To unsubscribe from this group, send email to > [email protected]<android-developers%[email protected]> > For more options, visit this group at > http://groups.google.com/group/android-developers?hl=en -- You received this message because you are subscribed to the Google Groups "Android Developers" 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-developers?hl=en
