如何根据呼叫的总持续时间从CallLog中获取前三个号码？
例如，给定这些调用/持续时间：
0123-2323 3分钟
0123-2323持续59分钟
3232-3210 15分钟0123-2323的总数为totalmin=62。对于3232-3210，它是totalmin=15。
我尝试了这段代码：
private String getCallDetails(){
StringBuffer sb = new StringBuffer();
Cursor managedCursor = managedQuery(CallLog.Calls.CONTENT_URI, null, null, null, null);
int number = managedCursor.getColumnIndex(CallLog.Calls.NUMBER);
int type = managedCursor.getColumnIndex(CallLog.Calls.TYPE);
int date = managedCursor.getColumnIndex(CallLog.Calls.DATE);
int duration = managedCursor.getColumnIndex(CallLog.Calls.DURATION);
sb.append("Call Details :");
while (managedCursor.moveToNext()){
String phNumber = managedCursor.getString(number);
String callType = managedCursor.getString(type);
String callDate = managedCursor.getString(date);
Date callDayTime = new Date(Long.valueOf(callDate));
String callDuration = managedCursor.getString(duration);
String dir = null;
int dircode = Integer.parseInt(callType);
switch (dircode) {
case CallLog.Calls.OUTGOING_TYPE:
dir = "OUTGOING";
break;
case CallLog.Calls.INCOMING_TYPE:
dir = "INCOMING";
break;
case CallLog.Calls.MISSED_TYPE:
dir = "MISSED";
break;
}
sb.append("\nPhone Number:--- " + phNumber + " \nCall Type:--- "
+ dir + " \nCall Date:--- " + callDayTime
+ " \nCall duration in sec :--- " + callDuration);
sb.append("\n----------------------------------");
}
managedCursor.close();
return sb.toString();
}
最佳答案
创建地图并对其进行排序。根据通话时间获取您的联系人，并像这样获得前三名联系人。
cc1=0;
Map<String, Integer> sortedMapa = sortByComparator(topsmstreemap, DESC);
for (Map.Entry<String, Integer> entry : sortedMapDeses.entrySet())
{
if(cc1<3)
{cc1++;
}