获取当前时间并检查时间是否已经过了一段时间

Question

以下代码获取该地区的当前时间和时区

    Date date = new Date();
    DateFormat df = new SimpleDateFormat("HH:mm:ss");

    df.setTimeZone(TimeZone.getDefault());

    System.out.println("Time: " + df.format(date)); 

现在是下午 1:01（打字时）

我需要帮助的是在代码中实现一个功能来检查当前时间是否已经过去，例如下午 1:00

但我什至不知道从哪里开始，你能帮我吗？

Answer 1

使用 Java 8+ Time API 类LocalTime：

LocalTime refTime = LocalTime.of(13, 0); // 1:00 PM

// Check if now > refTime, in default time zone
LocalTime now = LocalTime.now();
if (now.isAfter(refTime)) {
    // passed
}

// Check if now >= refTime, in pacific time zone
LocalTime now = LocalTime.now(ZoneId.of("America/Los_Angeles"))
if (now.compareTo(refTime) >= 0) {
    // passed
}

Answer 2

我看到它已经用 Time 回答了，但作为一个教学点，如果你真的想使用 Date，你可以这样做：

public static void main(String[] args) {
    Date date = new Date();
    DateFormat df = new SimpleDateFormat("HH:mm:ss");
    df.setTimeZone(TimeZone.getDefault());
    System.out.println("Time: " + df.format(date));

    //If you print the date you'll see how it is formatted
    //System.out.println(date.toString());

    //So you can just split the string and use the segment you want
    String[] fullDate = date.toString().split(" ");

    String compareAgainstTime = "01:00PM";

    System.out.println(isPastTime(fullDate[3],compareAgainstTime));
    }

public static boolean isPastTime(String currentTime, String comparedTime) {
    //We need to make the comparison time into the same format as the current time: 24H instead of 12H:
    //then we'll just convert the time into only minutes to that we can more easily compare;
    int comparedHour = comparedTime[-2].equals("AM") ? String.valueOf(comparedTime[0:2]) : String.valueOf(comparedTime[0:2] + 12 );
    int comparedMin = String.valueOf(comparedTime[3:5]);
    int comparedT = comparedHour*60 + comparedMin;

    //obviously currentTime is alredy the correct format; just need to convert to minutes
    int currentHour = String.valueOf(currentTime[0:2]);
    int currentMin = String.valueOf(currentTime[3:5]);
    int currentT = currentHour*60 + currentMin;

    return (currentT > comparedT);
}

这有点混乱，不得不混入弦乐之类的东西，但这是可能的。您还必须小心比较时间的零填充，或者只是在函数中检查它