Java 中的日期间隔加减法

Question

我正在 Java 中寻找一个库或助手 class，它可以让我执行日期间隔求和和减法。

例如，假设我有以下日期间隔：

A = ["2015-01-01 00:00", "2015-01-20 00:00"]
B = ["2015-01-05 00:00", "2015-01-10 00:00"]
C = ["2015-01-11 00:00", "2015-01-14 00:00"]
D = ["2015-01-19 00:00", "2015-01-25 00:00"]

1                  A               20
|----------------------------------|
    |---------|   |----------|   |------------|
    5    B    10  11    C    14  19    D      25

假设我想计算以下内容：

A - B - C + D = { ["2015-01-01 00:00", "2015-01-05 00:00"[,
                  ]"2015-01-10 00:00", "2015-01-11 00:00"[,
                  ]"2015-01-14 00:00", "2015-01-25 00:00"] }

1   5         10  11         14               25
|---|         |---|          |----------------|

我知道我可以使用纯 Java 构建自己的逻辑，但我不想重新发明轮子...

我正在研究 Joda-Time，但我不知道如何使用它执行此类操作。

非常感谢！

Answer 1

我认为基本上可以使用 Joda-Time 和一些自定义代码来完成。假定 A 是所有其他间隔应与之相关的间隔。

虽然这段代码应该给出预期的结果（并且应该相应地适用于不同的值），但我强烈建议使用非常不同的数据对其进行测试，尤其是对于三种情况 a) 一个完全不相交的区间 A , b) 在开头与 A 相交，c) 一个区间本身与 B 或 C 或 D.

相交

所以尽管如此，它可能有助于进一步的测试。

Interval a = new Interval(Instant.parse("2015-01-01T00:00Z"), Instant.parse("2015-01-20T00:00Z"));
List<Interval> l = Arrays.asList(
        /* b */ new Interval(Instant.parse("2015-01-05T00:00Z"), Instant.parse("2015-01-10T00:00Z")),
        /* c */ new Interval(Instant.parse("2015-01-11T00:00Z"), Instant.parse("2015-01-14T00:00Z")),
        /* d */ new Interval(Instant.parse("2015-01-19T00:00Z"), Instant.parse("2015-01-25T00:00Z"))
);
List<Interval> results = new ArrayList<Interval>();

for (Interval i : l) {
    if (a.contains(i)) {
        // if i is completely inside a, then calculate the first part and the remaining part
        //   whereas the first part will be added to the result
        Interval firstPart = new Interval(a.getStart(), i.getStart());
        results.add(firstPart);
        // followed by i itself (skipped)
        // part after i, inside a
        Interval remainingPart = new Interval(i.getEnd(), a.getEnd());
        a = remainingPart;
    } else if (i.overlaps(a)) {
        // if the intervals only overlap, then we take the earliest beginning and the latest ending as a result part
        DateTime overlapMin = (a.getStart().isBefore(i.getStart())) ? a.getStart() : i.getStart();
        DateTime overlapMax = (a.getEnd().isAfter(i.getEnd())) ? a.getEnd() : i.getEnd();
        Interval overlapAndBothParts = new Interval(overlapMin, overlapMax);
        results.add(overlapAndBothParts);
        // if the checked interval i is at the beginning, then a will become the part after this "overlap"
        if (i.getStartMillis() < a.getStartMillis()) {
            Interval whatsLeft = new Interval(i.getEndMillis(), a.getEndMillis());
            a = whatsLeft;
        }
    }
}

// print result
for (Interval i : results) {
    System.out.println("result part: " + i);
}

Answer 2

我找到了我需要的东西：Ranges, from the guava-libraries。

像这样工作：

Range<Date> a = Range.closed(
    new GregorianCalendar(2015, 0, 1).getTime(),
    new GregorianCalendar(2015, 0, 20).getTime());
Range<Date> b = Range.closed(
    new GregorianCalendar(2015, 0, 5).getTime(),
    new GregorianCalendar(2015, 0, 10).getTime());
Range<Date> c = Range.closed(
    new GregorianCalendar(2015, 0, 11).getTime(),
    new GregorianCalendar(2015, 0, 14).getTime());
Range<Date> d = Range.closed(
    new GregorianCalendar(2015, 0, 19).getTime(),
    new GregorianCalendar(2015, 0, 25).getTime());

RangeSet<Date> result = TreeRangeSet.create();
result.add(a);
result.remove(b);
result.remove(c);
result.add(d);

System.out.println(result);

上面的代码打印：

[
    [Thu Jan 01 00:00:00 BRST 2015‥Mon Jan 05 00:00:00 BRST 2015),
    (Sat Jan 10 00:00:00 BRST 2015‥Sun Jan 11 00:00:00 BRST 2015),
    (Wed Jan 14 00:00:00 BRST 2015‥Sun Jan 25 00:00:00 BRST 2015]
]

Java 中的日期间隔加减法

Date interval sum and subtraction in Java

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sum

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subtraction

intervals