计算2天android之间的时差

时间:2016-07-09 22:58:26

标签: android

我试图计算2次之间的差异。 它运作良好,除非两个日期在午夜的每一边。 在那种情况下,我得到一个否定的答案。 我有什么想法可以改进代码来解决这个问题吗?

String sleepStart = editFallAsleepTime.getText().toString();
                String sleepStop = editWakeUpTime.getText().toString();
                String awakeTimeString = ((Globals) getActivity().getApplication()).getAwakeTime();

//HH converts hour in 24 hours format (0-23), day calculation
                SimpleDateFormat format = new SimpleDateFormat("HH:mm");

                Date d1 = null;
                Date d2 = null;

                try
            {
                d1 = format.parse(sleepStart);
                d2 = format.parse(sleepStop);

                float t1 = d1.getTime();
                float t2 = d2.getTime();

                if( t2 <= t1 ){
                    t2 +=24;
                }

                float diff = t2 - t1;

                //in milliseconds
                    float diff = d2.getTime() - d1.getTime();
                    float diffHours = diff / (60 * 60 * 1000);
                    float awakeTime = Float.valueOf(awakeTimeString);
                    float awakeHours = awakeTime / 60;
                    float calcEffectiveSleep = diffHours - awakeHours;
                    String sleepTime = Float.toString(diffHours);
                    String effectiveSleep = Float.toString(calcEffectiveSleep);
                    ((Globals) getActivity().getApplication()).setEffectiveSleep(effectiveSleep);
                }
                catch (Exception e)
                {
                    Log.e("timediff","didntwork");
                }

1 个答案:

答案 0 :(得分:1)

如果您在1天间隔内的差异,那么您可以考虑处理d2&lt; d1案例正如@Andre Classen所说。怎么做是:

t1 = d1.getTime();
t2 = d2.getTime();

if( t2 <= t1 ){
   t2 +=24*60*60*1000;
}

diff = t2 - t1; 
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