3 
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());
4

2 回答 2

4

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（就我的谷歌技能而言）不存在。我已经以另一种您可能会觉得有用的方式解决了它。

在您的活动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它：

    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。

让我知道事情的后续。:)

于 2013-07-21T17:54:10.840 回答
1 
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);

    }

我对您提供的代码进行了一些修改，呵呵实际上在获取顶级联系人时存在问题，所以我在这里对其进行了修改：）顺便感谢您的帮助！

于 2013-07-26T07:52:49.650 回答