如果提前1、2、3、4分钟，如何将其向下舍入到最接近的5分钟间隔？

Question

我想将 Instant / LocalDateTime 舍入到 Java 中最接近的 5 分钟间隔。

示例：假设时间是：

2021-02-08T19:01:49.594 

或

2021-02-08T19:02:49.594

或

2021-02-08T19:03:49.594

或

2021-02-08T19:04:49.594

预期结果：

2021-02-08T19:00:00.000

Answer 1

为了获得更通用、可重用的解决方案，请实现自定义

TemporalUnit

与

truncatedTo​(TemporalUnit unit)

方法一起使用。以下是从另一个答案复制的实现。

然后您只需使用它以 5 分钟“单位”呼叫

truncatedTo()

。它适用于

LocalDateTime

和

Instant

。

String[] inputs = { "2021-02-08T19:01:49.594", "2021-02-08T19:02:49.594",
                    "2021-02-08T19:03:49.594", "2021-02-08T19:04:49.594",
                    "2021-02-08T19:26:49.594", "2021-02-08T19:27:49.594",
                    "2021-02-08T19:28:49.594", "2021-02-08T19:29:49.594" };
for (String input : inputs) {
    LocalDateTime dateTime = LocalDateTime.parse(input);
    Instant instant = Instant.parse(input + "Z");
    System.out.printf("%s -> %s   %s -> %s%n",
                      dateTime, dateTime.truncatedTo(DurationUnit.ofMinutes(5)),
                      instant, instant.truncatedTo(DurationUnit.ofMinutes(5)));
}

输出

2021-02-08T19:01:49.594 -> 2021-02-08T19:00   2021-02-08T19:01:49.594Z -> 2021-02-08T19:00:00Z
2021-02-08T19:02:49.594 -> 2021-02-08T19:00   2021-02-08T19:02:49.594Z -> 2021-02-08T19:00:00Z
2021-02-08T19:03:49.594 -> 2021-02-08T19:00   2021-02-08T19:03:49.594Z -> 2021-02-08T19:00:00Z
2021-02-08T19:04:49.594 -> 2021-02-08T19:00   2021-02-08T19:04:49.594Z -> 2021-02-08T19:00:00Z
2021-02-08T19:26:49.594 -> 2021-02-08T19:25   2021-02-08T19:26:49.594Z -> 2021-02-08T19:25:00Z
2021-02-08T19:27:49.594 -> 2021-02-08T19:25   2021-02-08T19:27:49.594Z -> 2021-02-08T19:25:00Z
2021-02-08T19:28:49.594 -> 2021-02-08T19:25   2021-02-08T19:28:49.594Z -> 2021-02-08T19:25:00Z
2021-02-08T19:29:49.594 -> 2021-02-08T19:25   2021-02-08T19:29:49.594Z -> 2021-02-08T19:25:00Z

作为通用用途，它实际上可以与

Instant

、

LocalDateTime

、

OffsetDateTime

、

ZonedDateTime

、

LocalTime

和

OffsetTime

一起使用，并且可以与划分为完整时间段的任何时间段一起使用天（例如 180 秒、或 5 分钟、或 2 小时，但不是 7 分钟）。

System.out.println(Instant.now().truncatedTo(DurationUnit.ofHours(2)));
System.out.println(LocalDateTime.now().truncatedTo(DurationUnit.ofHours(2)));
System.out.println(OffsetDateTime.now().truncatedTo(DurationUnit.ofHours(2)));
System.out.println(ZonedDateTime.now().truncatedTo(DurationUnit.ofHours(2)));
System.out.println(LocalTime.now().truncatedTo(DurationUnit.ofHours(2)));
System.out.println(OffsetTime.now().truncatedTo(DurationUnit.ofHours(2)));

输出

2021-02-15T06:00:00Z
2021-02-15T02:00
2021-02-15T02:00-05:00
2021-02-15T02:00-05:00[America/New_York]
02:00
02:00-05:00

自定义时间单位

public final class DurationUnit implements TemporalUnit {

    private static final int SECONDS_PER_DAY = 86400;
    private static final long NANOS_PER_SECOND =  1000_000_000L;
    private static final long NANOS_PER_DAY = NANOS_PER_SECOND * SECONDS_PER_DAY;

    private final Duration duration;

    public static DurationUnit of(Duration duration)   { return new DurationUnit(duration); }
    public static DurationUnit ofDays(long days)       { return new DurationUnit(Duration.ofDays(days)); }
    public static DurationUnit ofHours(long hours)     { return new DurationUnit(Duration.ofHours(hours)); }
    public static DurationUnit ofMinutes(long minutes) { return new DurationUnit(Duration.ofMinutes(minutes)); }
    public static DurationUnit ofSeconds(long seconds) { return new DurationUnit(Duration.ofSeconds(seconds)); }
    public static DurationUnit ofMillis(long millis)   { return new DurationUnit(Duration.ofMillis(millis)); }
    public static DurationUnit ofNanos(long nanos)     { return new DurationUnit(Duration.ofNanos(nanos)); }

    private DurationUnit(Duration duration) {
        if (duration.isZero() || duration.isNegative())
            throw new IllegalArgumentException("Duration may not be zero or negative");
        this.duration = duration;
    }

    @Override
    public Duration getDuration() {
        return this.duration;
    }

    @Override
    public boolean isDurationEstimated() {
        return (this.duration.getSeconds() >= SECONDS_PER_DAY);
    }

    @Override
    public boolean isDateBased() {
        return (this.duration.getNano() == 0 && this.duration.getSeconds() % SECONDS_PER_DAY == 0);
    }

    @Override
    public boolean isTimeBased() {
        return (this.duration.getSeconds() < SECONDS_PER_DAY && NANOS_PER_DAY % this.duration.toNanos() == 0);
    }

    @Override
    @SuppressWarnings("unchecked")
    public <R extends Temporal> R addTo(R temporal, long amount) {
        return (R) this.duration.multipliedBy(amount).addTo(temporal);
    }

    @Override
    public long between(Temporal temporal1Inclusive, Temporal temporal2Exclusive) {
        return Duration.between(temporal1Inclusive, temporal2Exclusive).dividedBy(this.duration);
    }

    @Override
    public String toString() {
        return this.duration.toString();
    }

}

Answer 2

您可以将其截断为

ChronoUnit.MINUTES

，然后根据要求检查分钟，即如果它不是

5

的倍数，则减去除以

5

时的余数。使用

LocalDate#withMinute

返回此

LocalDateTime

的副本，其中分钟已更改。

import java.time.LocalDateTime;
import java.time.format.DateTimeFormatter;
import java.time.temporal.ChronoUnit;

class Main {
    public static void main(String[] args) {
        // Test
        String[] arr = { "2021-02-08T19:02:49.594", "2021-02-08T19:56:49.594", "2021-02-08T19:54:49.594",
                "2021-02-08T19:06:49.594" };

        DateTimeFormatter dtf = DateTimeFormatter.ofPattern("uuuu-MM-dd'T'HH:mm:ss.SSS");

        for (String s : arr) {
            System.out.println(roundToNearestHour(s).format(dtf));
        }
    }

    static LocalDateTime roundToNearestHour(String str) {
        LocalDateTime ldt = LocalDateTime.parse(str).truncatedTo(ChronoUnit.MINUTES);
        int minute = ldt.getMinute();
        int remainder = minute % 5;
        if (remainder != 0) {
            ldt = ldt.withMinute(minute - remainder);
        }
        return ldt;
    }
}

输出：

2021-02-08T19:00:00.000
2021-02-08T19:55:00.000
2021-02-08T19:50:00.000
2021-02-08T19:05:00.000