如何获得Last Seen功能

Question

在过去的几天里，我尝试了最后一次在android中看到的功能，以便让人们活跃起来....我研究了许多答案，但我不满意...... 这是代码，将找到否。几天或几周或几年前这个人在线的人数...... 对于每个新的android编码器来说，这只是一块蛋糕......

GetTimeAgo getTimeAgo = new GetTimeAgo(); long lastTime = Long.parseLong(online); String lastSeenTime = getTimeAgo.getTimeAgo(lastTime, getApplicationContext()); mLastSeenView.setText(lastSeenTime);

通过上面的代码，你可以将它应用于任何活动......

打开.class文件GetTimeAgo 并应用以下代码...

public class GetTimeAgo extends Application {

    private static final int SECOND_MILLIS = 1000;
    private static final int MINUTE_MILLIS = 60 * SECOND_MILLIS;
    private static final int HOUR_MILLIS = 60 * MINUTE_MILLIS;
    private static final double DAY_MILLIS =  24 * HOUR_MILLIS;
    private static final double WEEK_MILLIS = 7 * DAY_MILLIS;
    private static final double MONTH_MILLIS = DAY_MILLIS * 30;
    private static final double YEAR_MILLIS = WEEK_MILLIS * 52;

    public static String getTimeAgo(long time, Context ctx) {
        if (time < 1000000000000L) {
            // if timestamp given in seconds, convert to millis
            time *= 1000;
        }

        long now = System.currentTimeMillis();
        if (time > now || time <= 0) {
            return null;
        }

        // TODO: localize
        final long diff = now - time;
        if (diff < MINUTE_MILLIS)
        {
            return "just now";
        }
        else if (diff < 2 * MINUTE_MILLIS)
        {
            return "a minute ago";
        }
        else if (diff < 50 * MINUTE_MILLIS)
        {

            double roundup = diff / MINUTE_MILLIS;
            int b = (int)(roundup);
            return b + " minutes ago";
        }
        else if (diff < 90 * MINUTE_MILLIS)
        {
            return "an hour ago";
        }
        else if (diff < 24 * HOUR_MILLIS)
        {
            double roundup = diff / HOUR_MILLIS;
            int b = (int)(roundup);
            return b + " hours ago";
        }
        else if (diff < 48 * HOUR_MILLIS)
        {
            return "yesterday";
        }
        else if (diff < 7 * DAY_MILLIS)
        {
            double roundup = diff / DAY_MILLIS;
            int b = (int)(roundup);
            return b + " days ago";
        }
        else if (diff < 2 * WEEK_MILLIS)
        {
            return "a week ago";
        }
        else if (diff < DAY_MILLIS * 30.43675)
        {
            double roundup = diff / WEEK_MILLIS;
            int b = (int)(roundup);
            return b + " weeks ago";
        }
        else if (diff < 2 * MONTH_MILLIS)
        {
            return "a month ago";
        }
        else if (diff < WEEK_MILLIS * 52.2)
        {
            double roundup = diff / MONTH_MILLIS;
            int b = (int)(roundup);
            return b + " months ago";
        }
        else if(diff < 2 * YEAR_MILLIS)
        {
            return "a year ago";
        }
        else
        {
            double roundup = diff / YEAR_MILLIS;
            int b = (int)(roundup);
            return b + " years ago";
        }
    }
}

如果你有更好的回答plz分享:)

Answer 1

我不确定你是否想要走这条路，但它可能有效。

使用Firebase，您可以执行此操作。

如果您在应用中打开实时数据库，则每次打开应用时都可以使用.setValue(value)设置数据库的日期，其中value类似于，

//Runs in your main loop

FirebaseDatabase database = FirebaseDatabase.getInstance();
DatabaseReference lastSeenData = database.getReference("message");

DateFormat dateFormat = new SimpleDateFormat("yyyy/MM/dd HH:mm:ss");
Date date = new Date();

lastSeenData.setValue(dateFormate.format(date));

您可能需要在.toString声明中致电setValue。

如果你想做类似＆＃34; So和So最后一次出现在2d之前＆＃34;，那么你可能必须得到日期So So So登录，然后从当前减去它日期。我的想法是，它更容易获得毫秒，用当前毫秒减去然后将差异作为格式化日期吐出。