android - 如何获取 callLog.calls 中最频繁的值条目的数量？

Question

Call Log Calls entry is this

 - 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

嗨，我想在 android 中查询，这样我只会检索Jed, 12345 << 因为他在列表中具有最重复的值，我想在 sqlite(android 查询)中执行此操作，但我不知道要调用哪些函数这是我使用的代码，但我只能获得最近调用的号码，而不是条目最多的号码。我如何进行查询？

    Date date=new Date() ;  

    Cursor c = contxt.getContentResolver().query(CallLog.Calls.CONTENT_URI,
            null, CallLog.Calls.TYPE + " AND " + CallLog.Calls.INCOMING_TYPE + 
            " AND " + 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());

Answer 1

如果您正在寻找一个查询来完成这项工作，我很遗憾地说它(就我的 google 技能而言)不存在。我已经用另一种您可能会觉得有用的方法解决了这个问题。

在您的 Activity 中的某处创建此函数:

public static void searchAndDisplay(ArrayList<String> arr) {

    ArrayList<String> list1 = new ArrayList();
    ArrayList<Integer> list2 = new ArrayList();
    for (int i = 0; i < 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);
        }
    }
    for (int i = 0; i < list1.size(); i++) {
        System.out.println("Number " + list1.get(i) + " occurs "
                + list2.get(i) + " times.");

    }
    int maxCount = 0;
    int index = -1;
    for (int i = 0; i < 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 you might want to do something/return the number with the highest occurences. 
}

然后在你想要光标的地方使用这个:

    Date date = new Date();
    ArrayList<String> allnumbers = new ArrayList();
    Cursor c = this.getContentResolver().query(
            CallLog.Calls.CONTENT_URI,
            null,
            CallLog.Calls.TYPE + " AND " + CallLog.Calls.INCOMING_TYPE
                    + " AND " + CallLog.Calls.DATE + ">=" + date.getDate(),
            null, CallLog.Calls.NUMBER);

    allnumbers.clear();
    if (c != null)
        c.moveToFirst();
    for (int i = 0; c.getCount() > i; i++) {

        String number1 = c.getString(0);

        allnumbers.add(number1);
        c.moveToNext();

    }
    searchAndDisplay(allnumbers);

您可能需要仔细检查您收到的号码是否正确。

让我知道进展如何。 :)