问题
I'm looking for a library or helper class in Java that would allow me to perform date interval sum and subtractions.
For example, lets's say I have the following date intervals:
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
And let's say I'd like to calculate the following:
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
|---| |---| |----------------|
I know I can build my own logic using pure Java, but I'd rather not reinvent the wheel...
I was looking into Joda-Time, but I couldn't figure out how to perform such operations using it.
Thanks a lot!
回答1:
I found exactly what I needed: Ranges, from the guava-libraries.
Works like this:
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);
The code above prints:
[
[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]
]
回答2:
I think it can be done basically using Joda-Time with some custom code. It is assumed that A
is the Interval which all other intervals should relate to.
While this code should give the expected results (and should work for different values accordingly) I highly suggest testing it with very different data, especially for the three cases a) an interval not intersecting A
at all, b) intersecting A
at the beginning and c) an interval which itself intersects B
or C
or D
.
So despite this, it might help for further tests.
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);
}
来源:https://stackoverflow.com/questions/28141773/date-interval-sum-and-subtraction-in-java