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 

或者

2021-02-08T19:02:49.594

或者

2021-02-08T19:03:49.594

或者

2021-02-08T19:04:49.594

预期结果：

2021-02-08T19:00:00.000

Answer 1

对于更通用、可重用的解决方案，请实现TemporalUnit与该方法一起使用的自定义truncatedTo\xe2\x80\x8b(TemporalUnit unit)。下面是从另一个答案复制的实现。

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然后，您只需使用它truncatedTo()以 5 分钟“单位”进行呼叫即可。它适用于LocalDateTime和Instant。

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

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输出

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2021-02-08T19:01:49.594 -> 2021-02-08T19:00   2021-02-08T19:01:49.594Z -> 2021-02-08T19:00:00Z\n2021-02-08T19:02:49.594 -> 2021-02-08T19:00   2021-02-08T19:02:49.594Z -> 2021-02-08T19:00:00Z\n2021-02-08T19:03:49.594 -> 2021-02-08T19:00   2021-02-08T19:03:49.594Z -> 2021-02-08T19:00:00Z\n2021-02-08T19:04:49.594 -> 2021-02-08T19:00   2021-02-08T19:04:49.594Z -> 2021-02-08T19:00:00Z\n2021-02-08T19:26:49.594 -> 2021-02-08T19:25   2021-02-08T19:26:49.594Z -> 2021-02-08T19:25:00Z\n2021-02-08T19:27:49.594 -> 2021-02-08T19:25   2021-02-08T19:27:49.594Z -> 2021-02-08T19:25:00Z\n2021-02-08T19:28:49.594 -> 2021-02-08T19:25   2021-02-08T19:28:49.594Z -> 2021-02-08T19:25:00Z\n2021-02-08T19:29:49.594 -> 2021-02-08T19:25   2021-02-08T19:29:49.594Z -> 2021-02-08T19:25:00Z\n

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Instant作为通用目的，它实际上可以与、LocalDateTime、OffsetDateTime、ZonedDateTime、LocalTime和一起使用OffsetTime，并且可以与分为一整天的任何时间段一起使用（例如 180 秒、或 5 分钟、或 2 小时，但不能是 7 分钟） 。

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

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输出

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2021-02-15T06:00:00Z\n2021-02-15T02:00\n2021-02-15T02:00-05:00\n2021-02-15T02:00-05:00[America/New_York]\n02:00\n02:00-05:00\n

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自定义时间单位

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public final class DurationUnit implements TemporalUnit {\n\n    private static final int SECONDS_PER_DAY = 86400;\n    private static final long NANOS_PER_SECOND =  1000_000_000L;\n    private static final long NANOS_PER_DAY = NANOS_PER_SECOND * SECONDS_PER_DAY;\n\n    private final Duration duration;\n\n    public static DurationUnit of(Duration duration)   { return new DurationUnit(duration); }\n    public static DurationUnit ofDays(long days)       { return new DurationUnit(Duration.ofDays(days)); }\n    public static DurationUnit ofHours(long hours)     { return new DurationUnit(Duration.ofHours(hours)); }\n    public static DurationUnit ofMinutes(long minutes) { return new DurationUnit(Duration.ofMinutes(minutes)); }\n    public static DurationUnit ofSeconds(long seconds) { return new DurationUnit(Duration.ofSeconds(seconds)); }\n    public static DurationUnit ofMillis(long millis)   { return new DurationUnit(Duration.ofMillis(millis)); }\n    public static DurationUnit ofNanos(long nanos)     { return new DurationUnit(Duration.ofNanos(nanos)); }\n\n    private DurationUnit(Duration duration) {\n        if (duration.isZero() || duration.isNegative())\n            throw new IllegalArgumentException("Duration may not be zero or negative");\n        this.duration = duration;\n    }\n\n    @Override\n    public Duration getDuration() {\n        return this.duration;\n    }\n\n    @Override\n    public boolean isDurationEstimated() {\n        return (this.duration.getSeconds() >= SECONDS_PER_DAY);\n    }\n\n    @Override\n    public boolean isDateBased() {\n        return (this.duration.getNano() == 0 && this.duration.getSeconds() % SECONDS_PER_DAY == 0);\n    }\n\n    @Override\n    public boolean isTimeBased() {\n        return (this.duration.getSeconds() < SECONDS_PER_DAY && NANOS_PER_DAY % this.duration.toNanos() == 0);\n    }\n\n    @Override\n    @SuppressWarnings("unchecked")\n    public <R extends Temporal> R addTo(R temporal, long amount) {\n        return (R) this.duration.multipliedBy(amount).addTo(temporal);\n    }\n\n    @Override\n    public long between(Temporal temporal1Inclusive, Temporal temporal2Exclusive) {\n        return Duration.between(temporal1Inclusive, temporal2Exclusive).dividedBy(this.duration);\n    }\n\n    @Override\n    public String toString() {\n        return this.duration.toString();\n    }\n\n}\n

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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除以 时减去余数的倍数5。用于返回更改了分钟的LocalDate#withMinute副本。LocalDateTime

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