本文整理汇总了Java中java.time.Duration.minusDays方法的典型用法代码示例。如果您正苦于以下问题：Java Duration.minusDays方法的具体用法？Java Duration.minusDays怎么用？Java Duration.minusDays使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类java.time.Duration的用法示例。
在下文中一共展示了Duration.minusDays方法的5个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Java代码示例。
示例1: formatAsElapsed
import java.time.Duration; //导入方法依赖的package包/类
public static String formatAsElapsed() {
Duration elapsed = Duration.ofMillis(ManagementFactory.getRuntimeMXBean().getUptime());
long days = elapsed.toDays();
elapsed = elapsed.minusDays(days);
long hours = elapsed.toHours();
elapsed = elapsed.minusHours(hours);
long minutes = elapsed.toMinutes();
elapsed = elapsed.minusMinutes(minutes);
long seconds = elapsed.getSeconds();
elapsed = elapsed.minusSeconds(seconds);
long millis = elapsed.toMillis();
return String.format(TIME_FORMAT,
days,
hours,
minutes,
seconds,
millis);
}
示例2: durationInMillisToString
import java.time.Duration; //导入方法依赖的package包/类
private static String durationInMillisToString(Duration duration) {
long days = duration.toDays(); // chop off any days as formatter can't handle them
duration = duration.minusDays(days);
// convert duration to a time by adding it to midnight
LocalTime durationAsTime = LocalTime.MIDNIGHT.plus(duration);
String result = formatter.format(durationAsTime);
if (days > 0) result = String.format("%dd ", days) + result;
return result;
}
示例3: minusDays_long
import java.time.Duration; //导入方法依赖的package包/类
@Test(dataProvider="MinusDays")
public void minusDays_long(long days, long amount, long expectedDays) {
Duration t = Duration.ofDays(days);
t = t.minusDays(amount);
assertEquals(t.toDays(), expectedDays);
}
示例4: minusDays_long_overflowTooBig
import java.time.Duration; //导入方法依赖的package包/类
@Test(expectedExceptions = {ArithmeticException.class})
public void minusDays_long_overflowTooBig() {
Duration t = Duration.ofDays(Long.MAX_VALUE/3600/24);
t.minusDays(-1);
}
示例5: minusDays_long_overflowTooSmall
import java.time.Duration; //导入方法依赖的package包/类
@Test(expectedExceptions = {ArithmeticException.class})
public void minusDays_long_overflowTooSmall() {
Duration t = Duration.ofDays(Long.MIN_VALUE/3600/24);
t.minusDays(1);
}