android - how to get the number of the most frequent value entry in callLog.calls? -

call log calls entry   - name        number         type          date called  - jed         12345          incoming      7-18-2013  - roger       14611          incoming      7-18-2013  - jed         12345          incoming      7-18-2013  - jed         12345          incoming      7-18-2013  - kevin       11111          incoming      7-18-2013 

hi, want query in android such retrieve jed, 12345 << since has repetitive value in list, im suppose in sqlite (android query) dont know functions invoke code used, able recent number called instead of 1 entries. how do query?

    date date=new date() ;        cursor c = contxt.getcontentresolver().query(calllog.calls.content_uri,             null, calllog.calls.type + " , " + calllog.calls.incoming_type +              " , " + calllog.calls.date + ">=" + date.getdate() ,             null,             calllog.calls.date + " desc limit 1");     if(c!=null)         do{             int callcounter = c.getcount();             string num = calllog_cursor.getstring(calllog_cursor                 .getcolumnindex(android.provider.calllog.calls.number));         }while(c.movetofirst()); 

if you're looking single query job, i'm sorry (as far sa google skills goes) doesn't exist. i've solved in way might find useful.

create function somewhere in activity:

public static void searchanddisplay(arraylist<string> arr) {      arraylist<string> list1 = new arraylist();     arraylist<integer> list2 = new arraylist();     (int = 0; < arr.size(); i++) {         int index = list1.indexof(arr.get(i));         if (index != -1) {             int newcount = list2.get(index) + 1;             list2.set(index, newcount);         } else {             list1.add(arr.get(i));             list2.add(1);         }     }     (int = 0; < list1.size(); i++) {         system.out.println("number " + list1.get(i) + " occurs "                 + list2.get(i) + " times.");      }     int maxcount = 0;     int index = -1;     (int = 0; < list2.size(); i++) {         if (maxcount < list2.get(i)) {             maxcount = list2.get(i);             index = i;         }     }     system.out.println("number " + arr.get(index)             + " has highest occurrence i.e " + maxcount); // here might want something/return number highest occurences.  } 

then want cursor use this:

    date date = new date();     arraylist<string> allnumbers = new arraylist();     cursor c = this.getcontentresolver().query(             calllog.calls.content_uri,             null,             calllog.calls.type + " , " + calllog.calls.incoming_type                     + " , " + calllog.calls.date + ">=" + date.getdate(),             null, calllog.calls.number);      allnumbers.clear();     if (c != null)         c.movetofirst();     (int = 0; c.getcount() > i; i++) {          string number1 = c.getstring(0);          allnumbers.add(number1);         c.movetonext();      }     searchanddisplay(allnumbers); 

you might want double-check numbers receive correct.

let me know how goes. :)