Java 获取前一天
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Get Previous Day
提问by Nathaniel Flath
Possible Duplicate:
How to determine the date one day prior to a given date in Java?
可能的重复：
如何在 Java 中确定给定日期前一天的日期？
If I have a Java.Util.Date object, what is the best way to get an object representing the 24 hours in the past of it?
如果我有一个 Java.Util.Date 对象，那么获取代表过去 24 小时的对象的最佳方法是什么？
采纳答案by Jared Oberhaus
Using Java 1.6 java.util.Calendar.add:
使用 Java 1.6 java.util.Calendar.add：
public static Date subtractDay(Date date) {
Calendar cal = Calendar.getInstance();
cal.setTime(date);
cal.add(Calendar.DAY_OF_MONTH, -1);
return cal.getTime();
}
Others suggest using Joda Time, which is currently JSR 310, and should later be included in Java itself.
回答by user101884
subtract 1000*60*60*24 from the time and create a new date.
从时间中减去 1000*60*60*24 并创建一个新日期。
Date yesterday = new Date(d.getTime() - (1000*60*60*24));
回答by Blake Pettersson
回答by Blake Pettersson
The important thing to remember is that the Date class should represent any points in time whilst the Calendar class is used to manipulate those points in time. Last of all, SimpleDateFormat will represent them as Strings.
需要记住的重要一点是 Date 类应该代表任何时间点，而 Calendar 类用于操作这些时间点。最后，SimpleDateFormat 将它们表示为字符串。
So, the best way is to use the Calendar class to calculate the new Date for you. This will ensure that any vagaries (Daylight Saving, Leap Years and the like) are accounted for.
因此，最好的方法是使用 Calendar 类为您计算新的日期。这将确保任何变幻莫测（夏令时、闰年等）都被考虑在内。
I'm assuming that you don't really want to find '24 Hours previous' but actually do want a new Date instance representing 'this time yesterday' - either way, you can ask the Calendar instance for a Date 24Hours prior to another or 1 Day prior.
我假设您并不是真的想找到“24 小时前”，但实际上确实想要一个新的日期实例来表示“昨天这个时间”-无论哪种方式，您都可以向日历实例询问另一个 24 小时之前的日期或1 天前。
The Daylight savings is a great example. The UK 'sprang forward' on the 26th March 2009. So, 1 day prior to 3.00a.m. on the 26.Mar.2009 should yield 3.00a.m. 25.Mar.2009 but 24 Hrs prior will yield 2.00a.m.
夏令时就是一个很好的例子。英国在 2009 年 3 月 26 日“向前冲”。因此，2009 年 3 月 26 日凌晨 3 点前 1 天应该产生 2009 年 3 月 25 日凌晨 3 点，但 24 小时前将产生 2 点
public class DateTests extends TestCase {
private static String EXPECTED_SUMMER_TIME = "2009.Mar.29 03:00:00";
private static String EXPECTED_SUMMER_TIME_LESS_DAY = "2009.Mar.28 03:00:00";
private static String EXPECTED_SUMMER_TIME_LESS_24_HRS = "2009.Mar.28 02:00:00";
private static String EXPECTED_SUMMER_TIME_LESS_FURTHER_24_HRS = "2009.Mar.27 02:00:00";
public void testSubtractDayOr24Hours() {
SimpleDateFormat formatter = new SimpleDateFormat("yyyy.MMM.dd HH:mm:SS");
Calendar calendar = Calendar.getInstance();
// Create our reference date, 3.00 a.m. on the day the clocks go forward (they 'went' forward at 02.00)
calendar.clear();
calendar.set(2009, 2, 29, 3, 0);
Date summerTime = calendar.getTime(); // Sun Mar 29 03:00:00 BST 2009
String formattedSummerTime = formatter.format(summerTime);
calendar.add(Calendar.DAY_OF_MONTH, -1);
// Our reference date less 'a day'
Date summerTimeLessADay = calendar.getTime(); // Sat Mar 28 03:00:00 GMT 2009
String formattedSummerTimeLessADay = formatter.format(summerTimeLessADay);
// reset the calendar instance to the reference day
calendar.setTime(summerTime);
// Our reference date less '24 hours' (is not quite 24 hours)
calendar.add(Calendar.HOUR, -24);
Date summerTimeLess24Hrs = calendar.getTime(); // Sat Mar 28 02:00:00 GMT 2009
String formattedSummerTimeLess24Hrs = formatter.format(summerTimeLess24Hrs);
// Third date shows that taking a further 24 hours from yields expected result
calendar.add(Calendar.HOUR, -24);
Date summerTimeLessFurther24Hrs = calendar.getTime(); // Fri Mar 27 02:00:00 GMT 2009
String formattedSummerTimeLessFurther24Hrs = formatter.format(summerTimeLessFurther24Hrs);
// reset the calendar once more to the day before
calendar.setTime(summerTimeLess24Hrs);
// Take a 'day' from the Sat will yield the same result as date 03 because Daylight Saving is not a factor
calendar.add(Calendar.DAY_OF_MONTH, -1);
Date summerTimeLessFurtherDay = calendar.getTime(); // Fri Mar 27 02:00:00 GMT 2009
String formattedSummerTimeLessFurtherDay = formatter.format(summerTimeLessFurtherDay);
assert(formattedSummerTime.equals(EXPECTED_SUMMER_TIME));
assert(formattedSummerTimeLessADay.equals(EXPECTED_SUMMER_TIME_LESS_DAY));
assert(formattedSummerTimeLess24Hrs.equals(EXPECTED_SUMMER_TIME_LESS_24_HRS));
assert(formattedSummerTimeLessFurther24Hrs.equals(EXPECTED_SUMMER_TIME_LESS_FURTHER_24_HRS));
// This last test proves that taking 24 hors vs. A Day usually yields the same result
assert(formattedSummerTimeLessFurther24Hrs.equals(formattedSummerTimeLessFurtherDay));
}
}
For testing date functions, wwwdot-timeanddate-dot-com is a great resource.
对于测试日期函数，wwwdot-timeanddate-dot-com 是一个很好的资源。
