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，我只会检索 Jed，12345＆lt;＆lt;因为他在列表中具有最重复的价值，我想在sqlite（android查询）中这样做但我不知道要调用哪些函数这是我使用的代码，但我只能得到最近调用的号码，而不是具有最多条目的号码。我怎么做这个问题？

    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

如果你正在寻找一个单独的查询来完成这项工作，我很遗憾地说它（就我的谷歌技能而言）并不存在。我用另一种你觉得有用的方式解决了它。

在您的活动中的某处创建此功能：

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. 
}

然后你想要光标的位置：

    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。

让我知道它是怎么回事。：）

Answer 2

public void getFrequentContact(ArrayList<String> logEntriesArray) {

    ArrayList<String> numArray        = new ArrayList<String>();
    ArrayList<Integer> callCountArray = new ArrayList<Integer>();
    int maxIndex = 0;

    for (int i = 0; i < logEntriesArray.size(); i++) {
        int index = numArray.indexOf(logEntriesArray.get(i));
        if (numArray.contains(logEntriesArray.get(i))) {
            int newCount = callCountArray.get(index) + 1;
            callCountArray.set(index, newCount);
        } else {
            numArray.add(logEntriesArray.get(i));
            callCountArray.add(1);
        }
    }


    for (int i = 0; i < numArray.size(); i++) {
        System.out.println("Number " + numArray.get(i) + " occurs "
                + callCountArray.get(i) + " times.");
    }


   if(callCountArray.size()>0){ 
     int maxValue = Collections.max(callCountArray);

        for(int i=0; i<callCountArray.size();i++){
            if(callCountArray.get(i)==maxValue)
                maxIndex = i;
        }
   }
    contactNumOfFreqCaller = numArray.get(maxIndex);
    Log.wtf(" FREQ NUM ", contactNumOfFreqCaller);

    }

我对你给出的代码做了一些修改实际上在获得顶级联系人方面存在问题，所以我对其进行了一些修改在这里它:)感谢帮助的方式!!

如何获取callLog.calls中最常值的条目数？

2 个答案: