i'm trying develop android app prevent user exit app using exit button.
i'm using broad cast receiver, in onreceive check if current running app app; if yes --> continue, else --> redirect app.
my question now: interval in receiver ( 3 second ) app takes 5-6 second re-open app, how can redirect app in less 3 second?
code:
manifest declaration:
<receiver android:name=".alarmreciever" /> implementation of receiver:
public class alarmreciever extends broadcastreceiver { @override public void onreceive(context context, intent intent) { activitymanager am1 = (activitymanager)context.getsystemservice(activity.activity_service); string packagename = am1.getrunningtasks(1).get(0).topactivity.getpackagename(); if (packagename.equals("com.xxxx.yyyyy")) { //continue; }else{ final intent intent = new intent(); intent.setflags(intent.flag_activity_new_task); intent.setclassname("com.xxxx.yyyyy", "com.xxxx.yyyyy.launch"); thisactivity.startactivity(intent); thisactivity.finish(); } } }
do want give provision exit own app? if that's case why have use broadcast receivers? why not try
android.os.process.killprocess(android.os.process.mypid());
onbackpressed() method?
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