Java 中的 Duration minusMinutes(long) 方法及示例

2025 年 1 月 6 日 | 阅读 6 分钟

Java中Duration类里的minusMinutes(long minutes)方法用于从一个Duration实例中减去指定分钟数。Duration类是java.time包中的一个类，它是一个在Java 8中引入的基于时间的功能，表示一段时间，例如，34.5秒。

方法签名

Duration: 方法的返回类型是Duration，这告诉我们该方法返回一个Duration类的对象。返回的Duration等于原始时长减去给定分钟数后的结果。

minusMinutes: 这是方法的名称。方法名本身就说明了该方法是为了从时长中减去几分钟而设计的。

long minutes: 该方法接受一个long类型的参数，名为minutes。该参数指定了要从调用该方法的Duration实例中减去的分钟数。

示例1：基本用法

在此示例中，我们使用ofHours方法创建一个表示2小时时长的Duration对象。该时长存储在duration变量中。我们在duration对象上调用minusMinutes(30)方法。这将从原始时长中减去30分钟，并返回一个新的Duration对象，我们将其存储在resultDuration变量中。我们打印原始时长以及减去30分钟后的时长。

文件名: DurationExample.java

import java.time.Duration;
public class DurationExample {
    public static void main(String[] args) {
        // Step 1: Create a Duration representing 2 hours
        Duration duration = Duration.ofHours(2);
        // Step 2: Subtract 30 minutes from the duration
        // Explanation: We use the minusMinutes method to subtract 30 minutes from the original duration.
        // The original duration remains unchanged as the Duration class is immutable.
        Duration resultDuration = duration.minusMinutes(30);
        // Step 3: Print the original and result durations
        // Explanation: We use System.out.println to display the durations to the console.
        // We concatenate the duration objects with strings for readability.
        System.out.println("Original Duration: " + duration);         // Output: PT2H
        System.out.println("Duration after subtracting 30 minutes: " + resultDuration); // Output: PT1H30M
    }
}

输出

Original Duration: PT2H
Duration after subtracting 30 minutes: PT1H30M

示例2：处理负结果

在此示例中，我们使用ofMinutes方法创建一个表示45分钟时长的Duration对象。我们尝试使用minusMinutes方法从原始时长中减去60分钟。由于减去的分钟数超过了原始时长，因此结果时长为负（PT-15M）。这表示一个过去15分钟的时长。我们打印原始时长（PT45M）和减去后的结果时长（PT-15M）作为输出。

文件名: DurationExample1.java

import java.time.Duration;
public class DurationExample1 {
    public static void main(String[] args) {
        // Example 1: Subtraction resulting in a negative duration
        // Create a Duration of 45 minutes
        Duration duration1 = Duration.ofMinutes(45);
        // Subtract 60 minutes from the duration
        Duration resultDuration1 = duration1.minusMinutes(60);
        // Print the original and result durations
        System.out.println("Example 1 - Original Duration: " + duration1);         // Output: PT45M
        System.out.println("Example 1 - Duration after subtracting 60 minutes: " + resultDuration1); // Output: PT-15M
        // Example 2: Subtraction resulting in a negative duration
        // Create a Duration of 1 hour
        Duration duration2 = Duration.ofHours(1);
        // Subtract 2 hours from the duration
        Duration resultDuration2 = duration2.minusMinutes(120);
        // Print the original and result durations
        System.out.println("Example 2 - Original Duration: " + duration2);         // Output: PT1H
        System.out.println("Example 2 - Duration after subtracting 2 hours: " + resultDuration2); // Output: PT-1H
    }
}

输出

Example 1 - Original Duration: PT45M
Example 1 - Duration after subtracting 60 minutes: PT-15M
Example 2 - Original Duration: PT1H
Example 2 - Duration after subtracting 2 hours: PT-1H

示例3：与其他方法链式调用

在此示例中，我们使用ofHours和plusMinutes方法创建了一个表示1小时20分钟时长的Duration对象。我们将minusMinutes(50)方法链式调用，以从原始时长中减去50分钟。然后，我们链式调用minusSeconds(10)方法，从先前获得的时长中进一步减去10秒。最终结果时长表示30分钟10秒。我们打印原始时长和结果时长的输出。

文件名: DurationExample2.java

import java.time.Duration;
public class DurationExample2 {
    public static void main(String[] args) {
        // Example 1: Basic Chaining
        // Create a Duration of 1 hour and 20 minutes
        Duration duration1 = Duration.ofHours(1).plusMinutes(20);
        // Subtract 50 minutes and 10 seconds from the duration
        Duration resultDuration1 = duration1.minusMinutes(50).minusSeconds(10);
        // Print the original and result durations
        System.out.println("Example 1 - Original Duration: " + duration1);
        System.out.println("Example 1 - Duration after subtracting 50 minutes and 10 seconds: " + resultDuration1);
        System.out.println(); // Adding a newline for clarity
        // Example 2: Chaining with Negative Result
        // Create a Duration of 1 hour
        Duration duration2 = Duration.ofHours(1);
        // Subtract 2 hours and 30 minutes from the duration
        Duration resultDuration2 = duration2.minusHours(2).minusMinutes(30);
        // Print the original and result durations
        System.out.println("Example 2 - Original Duration: " + duration2);
        System.out.println("Example 2 - Duration after subtracting 2 hours and 30 minutes: " + resultDuration2);
        System.out.println(); // Adding a newline for clarity
        // Example 3: Chaining with Large Numbers
        // Create a Duration of 5 hours
        Duration duration3 = Duration.ofHours(5);
        // Subtract a large number of minutes and seconds from the duration
        Duration resultDuration3 = duration3.minusMinutes(30000).minusSeconds(1000);
        // Print the original and result durations
        System.out.println("Example 3 - Original Duration: " + duration3);
        System.out.println("Example 3 - Duration after subtracting 30000 minutes and 1000 seconds: " + resultDuration3);
    }
}

输出

Example 1 - Original Duration: PT1H20M
Example 1 - Duration after subtracting 50 minutes and 10 seconds: PT29M50S
Example 2 - Original Duration: PT1H
Example 2 - Duration after subtracting 2 hours and 30 minutes: PT-1H-30M
Example 3 - Original Duration: PT5H
Example 3 - Duration after subtracting 30000 minutes and 1000 seconds: PT-495H-16M-40S

示例4：减去大量分钟数

在此示例中，我们创建了一个表示5小时时长的Duration对象。ofHours(5)方法初始化此时长并将其存储在duration变量中。我们使用minusMinutes(30000)从原始时长中减去30,000分钟。由于30,000分钟等于500小时（30,000 / 60），此减法导致大幅减少，使时长变为负数。然后，我们使用minusSeconds(1000)从已经为负的时长中减去1000秒。

文件名: LargeNumberOfMinutesExample.java

import java.time.Duration;
public class LargeNumberOfMinutesExample {
    public static void main(String[] args) {
        // Example 1: Subtracting a large number of minutes
        // Create a Duration of 5 hours
        Duration duration1 = Duration.ofHours(5);
        // Subtract 30,000 minutes from the duration
        Duration resultDuration1 = duration1.minusMinutes(30000);
        // Further subtract 1000 seconds from the duration
        resultDuration1 = resultDuration1.minusSeconds(1000);
        // Print the original and result durations
        System.out.println("Example 1 - Original Duration: " + duration1); // Output: PT5H
        System.out.println("Example 1 - Duration after subtracting 30,000 minutes and 1000 seconds: " + resultDuration1); // Output: PT-499H-1000S
        System.out.println(); // Adding a newline for clarity
        // Example 2: Subtracting enough minutes to span days
        // Create a Duration of 48 hours (2 days)
        Duration duration2 = Duration.ofDays(2);
        // Subtract 3 days worth of minutes (3 * 24 * 60 = 4320 minutes)
        Duration resultDuration2 = duration2.minusMinutes(4320);
        // Print the original and result durations
        System.out.println("Example 2 - Original Duration: " + duration2); // Output: PT48H
        System.out.println("Example 2 - Duration after subtracting 4320 minutes: " + resultDuration2); // Output: PT-24H
        System.out.println(); // Adding a newline for clarity
        // Example 3: Subtracting a large number of minutes from a shorter duration
        // Create a Duration of 1 hour
        Duration duration3 = Duration.ofHours(1);
        // Subtract 5000 minutes from the duration
        Duration resultDuration3 = duration3.minusMinutes(5000);
        // Further subtract 2000 seconds from the duration
        resultDuration3 = resultDuration3.minusSeconds(2000);
        // Print the original and result durations
        System.out.println("Example 3 - Original Duration: " + duration3); // Output: PT1H
        System.out.println("Example 3 - Duration after subtracting 5000 minutes and 2000 seconds: " + resultDuration3); // Output: PT-83H-2000S
    }
}

输出

Example 1 - Original Duration: PT5H
Example 1 - Duration after subtracting 30,000 minutes and 1000 seconds: PT-495H-16M-40S
Example 2 - Original Duration: PT48H
Example 2 - Duration after subtracting 4320 minutes: PT-24H
Example 3 - Original Duration: PT1H
Example 3 - Duration after subtracting 5000 minutes and 2000 seconds: PT-82H-53M-20S

应用

1. 调度系统

会议调度器：通过减去准备时间或缓冲时间来调整会议或活动时长。任务管理：当开始时间延迟或休息时，重新计算任务的剩余时间。

2. 事件倒计时器

倒计时时钟：从总时长中减去已过去的时间，以显示事件剩余时间。体育比赛计时器：根据暂停或中断动态调整比赛或对阵计时器。

3. 时间跟踪应用程序

工作日志：从总工作小时数中减去休息时间，以计算有效工作时间。项目管理：当项目任务发生延迟时，跟踪和调整项目任务的剩余时间。

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Java 中的 Duration minusMinutes(long) 方法及示例

方法签名

示例1：基本用法

示例2：处理负结果

示例3：与其他方法链式调用

示例4：减去大量分钟数

应用

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Java 中的 Duration minusMinutes(long) 方法及示例

方法签名

示例1：基本用法

示例2：处理负结果

示例3：与其他方法链式调用

示例4：减去大量分钟数

应用

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