本文整理了Java中org.threeten.bp.LocalTime.minusSeconds()
方法的一些代码示例,展示了LocalTime.minusSeconds()
的具体用法。这些代码示例主要来源于Github
/Stackoverflow
/Maven
等平台,是从一些精选项目中提取出来的代码,具有较强的参考意义,能在一定程度帮忙到你。LocalTime.minusSeconds()
方法的具体详情如下:
包路径:org.threeten.bp.LocalTime
类名称:LocalTime
方法名:minusSeconds
[英]Returns a copy of this LocalTime with the specified period in seconds subtracted.
This subtracts the specified number of seconds from this time, returning a new time. The calculation wraps around midnight.
This instance is immutable and unaffected by this method call.
[中]返回此LocalTime的副本,并减去指定的时间段(以秒为单位)。
这将从该时间减去指定的秒数,返回一个新时间。计算时间大约在午夜。
此实例是不可变的,不受此方法调用的影响。
代码示例来源:origin: stackoverflow.com
LocalTime time1 = LocalTime.of(13, 11);
LocalTime untilTime = time1.minusSeconds(60);
LocalTime currentTime = LocalTime.now();
Duration duration = Duration.between(currentTime, untilTime);
// TODO possibly check duration.isNegative()
Thread.sleep(duration.toMillis());
代码示例来源:origin: ThreeTen/threetenbp
/**
* Returns a copy of this {@code OffsetTime} with the specified period in seconds subtracted.
* <p>
* This subtracts the specified number of seconds from this time, returning a new time.
* The calculation wraps around midnight.
* <p>
* This instance is immutable and unaffected by this method call.
*
* @param seconds the seconds to subtract, may be negative
* @return an {@code OffsetTime} based on this time with the seconds subtracted, not null
*/
public OffsetTime minusSeconds(long seconds) {
return with(time.minusSeconds(seconds), offset);
}
代码示例来源:origin: org.threeten/threetenbp
/**
* Returns a copy of this {@code OffsetTime} with the specified period in seconds subtracted.
* <p>
* This subtracts the specified number of seconds from this time, returning a new time.
* The calculation wraps around midnight.
* <p>
* This instance is immutable and unaffected by this method call.
*
* @param seconds the seconds to subtract, may be negative
* @return an {@code OffsetTime} based on this time with the seconds subtracted, not null
*/
public OffsetTime minusSeconds(long seconds) {
return with(time.minusSeconds(seconds), offset);
}
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