我在这里看到了所有的帖子，但我仍然无法弄清楚如何在两个安卓日期之间找到差异。
这就是我的所作所为：
long diff = date1.getTime() - date2.getTime();
Date diffDate = new Date(diff);
我得到：日期是1970年1月1日，时间总是在两个小时内变大......我来自以色列所以两个小时是时间偏离。
我怎样才能得到正常的差异？
答案 0 :(得分：73)
你接近正确的答案，你得到这两个日期之间的毫秒差异，但是当你试图用这个差异构建一个日期时，假设你要创建一个新的{{1}具有该差异值的对象作为其纪元时间。如果你正在寻找一个小时的时间，那么你只需要对Date做一些基本的算术来得到不同的时间部分：
diff所有这些数学运算只会进行整数运算，因此会截断任何小数点
答案 1 :(得分：22)
long diffInMillisec = date1.getTime() - date2.getTime();
long diffInDays = TimeUnit.MILLISECONDS.toDays(diffInMillisec);
long diffInHours = TimeUnit.MILLISECONDS.toHours(diffInMillisec);
long diffInMin = TimeUnit.MILLISECONDS.toMinutes(diffInMillisec);
long diffInSec = TimeUnit.MILLISECONDS.toSeconds(diffInMillisec);
答案 2 :(得分：14)
一些补充： 在这里我将字符串转换为日期，然后我比较当前时间。
String toyBornTime = "2014-06-18 12:56:50";
SimpleDateFormat dateFormat = new SimpleDateFormat(
"yyyy-MM-dd HH:mm:ss");
try {
Date oldDate = dateFormat.parse(toyBornTime);
System.out.println(oldDate);
Date currentDate = new Date();
long diff = currentDate.getTime() - oldDate.getTime();
long seconds = diff / 1000;
long minutes = seconds / 60;
long hours = minutes / 60;
long days = hours / 24;
if (oldDate.before(currentDate)) {
Log.e("oldDate", "is previous date");
Log.e("Difference: ", " seconds: " + seconds + " minutes: " + minutes
+ " hours: " + hours + " days: " + days);
}
// Log.e("toyBornTime", "" + toyBornTime);
} catch (ParseException e) {
e.printStackTrace();
}
答案 3 :(得分：2)
使用java.time.Duration：
Duration diff = Duration.between(instant2, instant1);
System.out.println(diff);
这将打印类似
的内容PT109H27M21S
这意味着一段时间为109小时27分21秒。如果你想要更具人性化的东西 - 我会先给出9版本，这是最简单的：
String formattedDiff = String.format(Locale.ENGLISH,
"%d days %d hours %d minutes %d seconds",
diff.toDays(), diff.toHoursPart(), diff.toMinutesPart(), diff.toSecondsPart());
System.out.println(formattedDiff);
现在我们得到
4 days 13 hours 27 minutes 21 seconds
Duration类是现代Java日期和时间API java.time的一部分。这是捆绑在较新的Android设备上。在旧设备上，获取ThreeTenABP并将其添加到项目中，并确保从同一个包中导入org.threeten.bp.Duration和其他日期时间类。
假设您还没有Java 9版本，您可以依次减去较大的单位以获得较小的单位：
long days = diff.toDays();
diff = diff.minusDays(days);
long hours = diff.toHours();
diff = diff.minusHours(hours);
long minutes = diff.toMinutes();
diff = diff.minusMinutes(minutes);
long seconds = diff.toSeconds();
然后你可以格式化上面的四个变量。
Date表示某个时间点。它从来没有意味着代表一定的时间，持续时间，并且它不适合它。尝试完成这项工作充其量会导致令人困惑且难以维护的代码。你不希望如此，所以请不要。
java.time。java.time向Java 6和7 答案 4 :(得分：2)
如果您使用Kotlin语言进行Android开发，则可以使用ExperimentalTime扩展名。要获得不同的日子，可以这样使用它：
@ExperimentalTime
fun daysDiff(c1: Calendar, c2: Calendar): Double {
val diffInMillis = c1.timeInMillis - c2.timeInMillis
return diffInMillis.milliseconds.inDays
}
或者如果要获取整数形式的结果：
@ExperimentalTime
fun daysDiff2(c1: Calendar, c2: Calendar): Int {
val diffInMillis = c1.timeInMillis - c2.timeInMillis
return diffInMillis.milliseconds.toInt(DurationUnit.DAYS)
}
答案 5 :(得分：1)
这是我基于@Ole V. V.答案的答案。
这也适用于Singular。
private String getDuration(Date d1, Date d2) {
Duration diff = Duration.between(d1.toInstant(), d2.toInstant());
long days = diff.toDays();
diff = diff.minusDays(days);
long hours = diff.toHours();
diff = diff.minusHours(hours);
long minutes = diff.toMinutes();
diff = diff.minusMinutes(minutes);
long seconds = diff.toMillis();
StringBuilder formattedDiff = new StringBuilder();
if(days!=0){
if(days==1){
formattedDiff.append(days + " Day ");
}else {
formattedDiff.append(days + " Days ");
}
}if(hours!=0){
if(hours==1){
formattedDiff.append(hours + " hour ");
}else{
formattedDiff.append(hours + " hours ");
}
}if(minutes!=0){
if(minutes==1){
formattedDiff.append(minutes + " minute ");
}else{
formattedDiff.append(minutes + " minutes ");
}
}if(seconds!=0){
if(seconds==1){
formattedDiff.append(seconds + " second ");
}else{
formattedDiff.append(seconds + " seconds ");
}
}
return formattedDiff.toString();
}
它与StringBuilder一起将所有内容附加在一起。
答案 6 :(得分：1)
我尝试过这种方法..但不知道为什么我没有得到正确的结果
long diff = date1.getTime() - date2.getTime();
long seconds = diff / 1000;
long minutes = seconds / 60;
long hours = minutes / 60;
long days = hours / 24;
但这可行
long miliSeconds = date1.getTime() -date2.getTime();
long seconds = TimeUnit.MILLISECONDS.toSeconds(miliSeconds);
long minute = seconds/60;
long hour = minute/60;
long days = hour/24;
答案 7 :(得分：1)
如何使用 Instant：
annotations:
kubed.appscode.com/sync: "cert-manager-tls=dev"
您甚至可以使用 instant1.toString() 保存您的 Instant 并使用 parse(string) 解析该字符串。
如果您需要支持 Android API 级别 < 26 只需添加 Java 8+ API desugaring support 到您的项目。
答案 8 :(得分：0)
使用这些功能
public static int getDateDifference(
int previousYear, int previousMonthOfYear, int previousDayOfMonth,
int nextYear, int nextMonthOfYear, int nextDayOfMonth,
int differenceToCount){
// int differenceToCount = can be any of the following
// Calendar.MILLISECOND;
// Calendar.SECOND;
// Calendar.MINUTE;
// Calendar.HOUR;
// Calendar.DAY_OF_MONTH;
// Calendar.MONTH;
// Calendar.YEAR;
// Calendar.----
Calendar previousDate = Calendar.getInstance();
previousDate.set(Calendar.DAY_OF_MONTH, previousDayOfMonth);
// month is zero indexed so month should be minus 1
previousDate.set(Calendar.MONTH, previousMonthOfYear);
previousDate.set(Calendar.YEAR, previousYear);
Calendar nextDate = Calendar.getInstance();
nextDate.set(Calendar.DAY_OF_MONTH, previousDayOfMonth);
// month is zero indexed so month should be minus 1
nextDate.set(Calendar.MONTH, previousMonthOfYear);
nextDate.set(Calendar.YEAR, previousYear);
return getDateDifference(previousDate,nextDate,differenceToCount);
}
public static int getDateDifference(Calendar previousDate,Calendar nextDate,int differenceToCount){
// int differenceToCount = can be any of the following
// Calendar.MILLISECOND;
// Calendar.SECOND;
// Calendar.MINUTE;
// Calendar.HOUR;
// Calendar.DAY_OF_MONTH;
// Calendar.MONTH;
// Calendar.YEAR;
// Calendar.----
//raise an exception if previous is greater than nextdate.
if(previousDate.compareTo(nextDate)>0){
throw new RuntimeException("Previous Date is later than Nextdate");
}
int difference=0;
while(previousDate.compareTo(nextDate)<=0){
difference++;
previousDate.add(differenceToCount,1);
}
return difference;
}
答案 9 :(得分：0)
使用格鲁吉亚卡兰德
public void dateDifferenceExample() {
// Set the date for both of the calendar instance
GregorianCalendar calDate = new GregorianCalendar(2012, 10, 02,5,23,43);
GregorianCalendar cal2 = new GregorianCalendar(2015, 04, 02);
// Get the represented date in milliseconds
long millis1 = calDate.getTimeInMillis();
long millis2 = cal2.getTimeInMillis();
// Calculate difference in milliseconds
long diff = millis2 - millis1;
// Calculate difference in seconds
long diffSeconds = diff / 1000;
// Calculate difference in minutes
long diffMinutes = diff / (60 * 1000);
// Calculate difference in hours
long diffHours = diff / (60 * 60 * 1000);
// Calculate difference in days
long diffDays = diff / (24 * 60 * 60 * 1000);
Toast.makeText(getContext(), ""+diffSeconds, Toast.LENGTH_SHORT).show();
}
答案 10 :(得分：0)
用Kotlin编写： 如果您需要两个日期之间的差额，并且不关心日期本身（例如，如果您需要在应用中执行某些操作，这是基于其他操作时间（例如保存在共享首选项中的时间），那么很好）。 第一次保存：
val firstTime:Long= System.currentTimeMillis()
第二次保存：
val now:Long= System.currentTimeMillis()
计算两次之间的毫秒数：
val milisecondsSinceLastTime: Long =(now-lastScrollTime)
答案 11 :(得分：-1)
最短的答案对我有用。以毫秒为单位发送开始和结束日期。
public int GetDifference(long start,long end){
Calendar cal = Calendar.getInstance();
cal.setTimeInMillis(start);
int hour = cal.get(Calendar.HOUR_OF_DAY);
int min = cal.get(Calendar.MINUTE);
long t=(23-hour)*3600000+(59-min)*60000;
t=start+t;
int diff=0;
if(end>t){
diff=(int)((end-t)/ TimeUnit.DAYS.toMillis(1))+1;
}
return diff;
}