如何让ArrayList中重复的联系人数量 [英] how to get count of duplicate contacts in arraylist

问题描述

我有一个包含从通话记录编号，名称的ArrayList。从此ArrayList我用哪个先检查ArrayList中的数量比返回结果的方法删除重复的。结果
 我的code是

I have an Arraylist which contains Number,Name from call log. to remove the duplicate from this arraylist I used a method Which first check the number in the arraylist than return the result.
My code is

public void getnumber(ContentResolver cr) {
    aa.clear();
    num.clear();
    String viaOrder = android.provider.CallLog.Calls.DATE + " DESC," + android.provider.CallLog.Calls.NUMBER + " DESC";
    Cursor c = getContentResolver().query(CallLog.Calls.CONTENT_URI, null,
            null,
            null,
            viaOrder);
    while (c.moveToNext()) {
        Info info = new Info();
        info.phone=c.getString(c.getColumnIndex(android.provider.CallLog.Calls.NUMBER));
        if(num.contains(info.phone)){
            count++;

            //int cnt = aa.indexOf(count);
            //aa.set(cnt, count);
        }
        else {
            num.add(info.phone);
            String nam =c.getString(c.getColumnIndex(android.provider.CallLog.Calls.CACHED_NAME));
            if(nam==null){
                nam = "Unknown";
            }
            nam+=count;
            byte[] imagearray = null;
            aa.add(new ContactStock(nam,info.phone,imagearray,count));
            count=1;
        }   
      }
    c.close();
}

这一切工作正常，现在我得每个联系人的数量。这样我可以安排通过最联络的联系人名单。我尝试了很多，但它给了我错误的输出。任何帮助将AP preciated。

All this working fine Now I have to get count of each contact. so that I can arrange my list by most contacted contacts. I tried it a lot but it gives me wrong output. Any help Will be appreciated.

推荐答案

试试这个，

 public void getnumber(ContentResolver cr) {
    aa.clear();
    num.clear();
    String viaOrder = android.provider.CallLog.Calls.DATE + " DESC," + android.provider.CallLog.Calls.NUMBER + " DESC";
    Cursor c = getContentResolver().query(CallLog.Calls.CONTENT_URI, null,
            null,
            null,
            viaOrder);
    while (c.moveToNext()) {
        Info info = new Info();
        info.phone=c.getString(c.getColumnIndex(android.provider.CallLog.Calls.NUMBER));
        info.name = c.getString(c.getColumnIndex(android.provider.CallLog.Calls.CACHED_NAME));
        num.add(info.toString());
      }
    c.close();
    Map<String, Integer> map = new LinkedHashMap<String, Integer>();

    for (String temp : num) {
        Integer count = map.get(temp);
        map.put(temp, (count == null) ? 1 : count + 1);
    }
    printMap(map);

    //System.out.println("\nSorted Map");
    Map<String, Integer> treeMap = new TreeMap<String, Integer>(map);
    printMap(treeMap);
}
public static void printMap(Map<String, Integer> map){

    for (Map.Entry<String, Integer> entry : map.entrySet()) {
        System.out.println("Key : " + entry.getKey() + " Value : "
            + entry.getValue());
        aa.add(new ContactStock("Name "+entry.getValue(),entry.getKey() , imagearray, entry.getValue()));
    }
}

LinkedHashset保留O对你的列表中的顺序。 这里指了解更多详情

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