how to close a webview which was opened by javascript in android programming -


i have android application contains 1 web view , have loaded url(which developed .net using javascript)in web view , has facebook login , when user click on facebook button,it opens facebook dialog,if user don't want login,he click on go on android facebook dialog still appears,how go fom facebook dialog main web view?

i tried code

 @override public boolean onkeydown(int keycode, keyevent event) {     if (event.getaction() == keyevent.action_down) {         switch (keycode) {             case keyevent.keycode_back:                 if (mainwebview.cangoback()) {                     mainwebview.goback();                     log.d("webview<---", "going_back");                 } else if (newwebview != null && newwebview.isfocusable()) {                     fbwebview.removeallviews();                     fbwebview.clearhistory();                     fbwebview.getparent();                        fbwebview=null;                     //fbwebview.addview(mwebview);                     // fbwebview.loadurl(url);                     //startactivity(new intent(mainactivity.this, mainactivity.class).addflags(intent.flag_activity_new_task).addflags(intent.flag_activity_clear_top));                     log.d("fb_webview<---", "finish");                  }                 return true;         }      }     return super.onkeydown(keycode, event); }

still no use,can suggest how go fb dialog main web view ,here fbwebview created dynamically in activity , mainwebview intialised through xml , how close fbwebview appear mainwebview?

write below code in before return statement under keyevent.keycode_back case:

  /* closing newly created popup window */                 if (build.version.sdk_int < 18) {                     newwebview.clearview();                 } else {                     newwebview.loadurl("about:blank");                 }

Comments

Post a Comment

Popular posts from this blog

python - Operations inside variables -

this question has answer here: evaluating mathematical expression in string 12 answers i want make python script finds maximum possible output using 2 operations. problem says there syntax error @ 2nd last line, in b in there. how can fix it? x = 0 = int(input("1st number:")) c = int(input("2nd number:" )) e = int(input("3rd number:" )) in range(4): if == 0: b = "+" elif == 1: b = "-" elif == 2: b = "/" else: b = "*" j in range(4): if j == 0: d = "+" elif j == 1: d = "-" elif j == 2: d = "/" else: d = "*" k = b c d e print(k) if you're after variables e on single line, couldn...
Read more

Generic Map Parameter java -

i having lot of methods of type: public static list<eduusers> getdetails(class c, map<string, integer> params) { list<eduusers> details = null; session session = hibernateutil.getsessionfactory().opensession(); transaction tx = null; try { tx = session.begintransaction(); criteria cr = session.createcriteria(c); (map.entry<string, integer> entry : params.entryset()) { cr.add(restrictions.eq(entry.getkey(), entry.getvalue())); } details = cr.list(); tx.commit(); } catch (exception asd) { log.debug(asd.getmessage()); if (tx != null) { tx.commit(); } } { session.close(); } return details; } i trying have generic method them , have written far: public static <t> list<t> getdetails(class<t> c, class<t> b) { l...
Read more

arrays - What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it? -

what arrayindexoutofboundsexception mean , how rid of it? here code sample triggers exception: string[] name = {"tom", "dick", "harry"}; for(int = 0; i<=name.length; i++) { system.out.print(name[i] +'\n'); } your first port of call should documentation explains reasonably clearly: thrown indicate array has been accessed illegal index. index either negative or greater or equal size of array. so example: int[] array = new int[5]; int boom = array[10]; // throws exception as how avoid it... um, don't that. careful array indexes. one problem people run thinking arrays 1-indexed, e.g. int[] array = new int[5]; // ... populate array here ... (int index = 1; index <= array.length; index++) { system.out.println(array[index]); } that miss out first element (index 0) , throw exception when index 5. valid indexes here 0-4 inclusive. correct, idiomatic for statement here be: for (int index = 0; index <...
Read more