Question

我正在开发Android应用程序，我需要计算一些时间（使用Joda libs）。我想知道是否存在这样做的Java类。

我想计算的案例（例如）：

案例1：

t1 = "13:40"
t2 = "12:30"
result = t1 + t2 --> 26:10

案例2：

t1 = "13:40"
t2 = "12:30"
result = t1 - t2 --> 1:10

案例2.1：

t1 = "13:40"
t2 = "14:00"
result = t1 - t2 --> -0:20

谢谢！：）

Answer 1

以下JodaTime-code处理输入和结果作为时间量，而不是时间点：

案例1：

PeriodFormatterBuilder builder = new PeriodFormatterBuilder();
builder.minimumPrintedDigits(2);
builder.printZeroAlways();
builder.appendHours();
builder.appendLiteral(":");
builder.appendMinutes();
PeriodFormatter pf = builder.toFormatter();

Period period1 = pf.parsePeriod("13:40");
Period period2 = pf.parsePeriod("12:30");
Period total = period1.plus(period2);
Period normalized = total.normalizedStandard(PeriodType.time());

System.out.println("TOTAL PERIOD: " + total); // PT25H70M
System.out.println("NORMALIZED PERIOD: " + normalized); // PT26H10M
System.out.println("FORMATTED NORMALIZED PERIOD: " + pf.print(normalized)); // 26:10

案例2 + 2.1：

PeriodFormatterBuilder builder = new PeriodFormatterBuilder();
builder.minimumPrintedDigits(2);
builder.printZeroAlways();
builder.appendHours();
builder.appendLiteral(":");
builder.appendMinutes();
PeriodFormatter pf = builder.toFormatter();

Period period1 = pf.parsePeriod("13:40");
Period period2 = pf.parsePeriod("14:00");

boolean negative = false;

if (period1.toStandardDuration().isShorterThan(period2.toStandardDuration())) {
  Period tmp = period1;
  period1 = period2;
  period2 = tmp;
  negative = true;
}

Period total = period1.minus(period2);
Period normalized = total.normalizedStandard(PeriodType.time());
System.out.println("NORMALIZED PERIOD: " + (negative ? "-" : "") + normalized);
System.out.println(
  "FORMATTED NORMALIZED PERIOD: " + (negative ? "-" : "") + pf.print(normalized)
); // -00:20

请注意，由于JodaTime的标志处理不充分（喜欢打印不在前面的标志，而是在句点表达式中打印宽度），因此需要一些笨重的解决方法。

Answer 2

我在这里计算05:00:00 PM这种格式你可以采取任何你想要的格式：

    String string1 = "05:00:00 PM";
    Date time1 = new SimpleDateFormat("HH:mm:ss aa").parse(string1);
    Calendar calendar1 = Calendar.getInstance();
    calendar1.setTime(time1);

    String string2 = "09:00:00 AM";
    Date time2 = new SimpleDateFormat("HH:mm:ss aa").parse(string2);
    Calendar calendar2 = Calendar.getInstance();
    calendar2.setTime(time2);
    calendar2.add(Calendar.DATE, 1);

    Date x = calendar1.getTime();
    Date xy = calendar2.getTime();
    long diff = x.getTime() - xy.getTime();
    diffMinutes = diff / (60 * 1000);
    float diffHours = diffMinutes / 60;
    System.out.println("diff hours" + diffHours);

Java中的小时数和时间之差（超过24小时）

2 个答案: