Java Instant minusSeconds()方法及示例

来源:这里教程网 时间:2026-02-17 20:51:54 作者:

Java Instant minusSeconds()方法及示例

Instant类minusSeconds() 方法从这个瞬间减去指定的第二个值,并将结果作为一个瞬间对象返回。这个瞬时是不可改变的。
语法:

public Instant minusSeconds(long secondsToSubtract)

参数: 该方法接受一个参数 secondsToSubtract ,即要减去的秒数。

返回: 该方法返回减去秒数后的 Instant

异常。该方法会抛出以下异常:

DateTimeException :如果结果超过了最大或最小的瞬间。ArithmeticException :如果发生数字溢出。

以下程序说明了minusSeconds()方法:

程序1:

// Java program to demonstrate// Instant.minusSeconds() method import java.time.*; public class GFG {    public static void main(String[] args)    {         // create a Instant object        Instant instant            = Instant.parse("2018-10-30T09:05:55.13Z");         // current Instant        System.out.println("Initialize instant: "                           + instant);         // subtract 4300 seconds from this instant        Instant returnedValue            = instant.minusSeconds(4300);         // print result        System.out.println("Returned Instant: "                           + returnedValue);    }}

输出

Initialize instant: 2018-10-30T09:05:55.130ZReturned Instant: 2018-10-30T07:54:15.130Z

程序2:

// Java program to demonstrate// Instant.minusSeconds() method import java.time.*; public class GFG {    public static void main(String[] args)    {         // create a Instant object        Instant instant = Instant.now();         // current Instant        System.out.println("Current instant: "                           + instant);         // subtract 54000 from this instant        Instant returnedValue            = instant.minusSeconds(54000);         // print result        System.out.println("Returned Instant: "                           + returnedValue);    }}

输出

Current instant: 2018-11-27T06:44:04.901ZReturned Instant: 2018-11-26T15:44:04.901Z

参考文献: https://docs.oracle.com/javase/10/docs/api/java/time/Instant.html#minusSeconds(long)

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