无论日期如何，检查给定时间是否介于两次之间

2022-01-11 00:00:00 date calendar android java timespan

我有时间跨度:

字符串时间1 = 01:00:00

String time1 = 01:00:00

字符串时间2 = 05:00:00

String time2 = 05:00:00

我想检查 time1 和 time2 是否都在 20:11:13 和 14:49:00 之间.

I want to check if time1 and time2 both lies between 20:11:13 and 14:49:00.

实际上，考虑到，01:00:00大于20:11:13且小于14:49:00>20:11:13 始终小于 14:49:00.这是先决条件.

Actually, 01:00:00 is greater than 20:11:13 and less than 14:49:00 considering 20:11:13 is always less than 14:49:00. This is given prerequisite.

所以我想要的是，20:11:13 <01:00:00 <14:49:00.

所以我需要这样的东西:

So I need something like that:

 public void getTimeSpans()
{
    boolean firstTime = false, secondTime = false;
    
    if(time1 > "20:11:13" && time1 < "14:49:00")
    {
       firstTime = true;
    }
    
    if(time2 > "20:11:13" && time2 < "14:49:00")
    {
       secondTime = true;
    }
 }

我知道这段代码在比较字符串对象时没有给出正确的结果.

I know that this code does not give correct result as I am comparing the string objects.

如何做到这一点，因为它们是时间跨度而不是要比较的字符串?

How to do that as they are the timespans but not the strings to compare?

推荐答案

您可以使用 Calendar 类进行检查.

You can use the Calendar class in order to check.

例如:

try {
    String string1 = "20:11:13";
    Date time1 = new SimpleDateFormat("HH:mm:ss").parse(string1);
    Calendar calendar1 = Calendar.getInstance();
    calendar1.setTime(time1);
    calendar1.add(Calendar.DATE, 1);


    String string2 = "14:49:00";
    Date time2 = new SimpleDateFormat("HH:mm:ss").parse(string2);
    Calendar calendar2 = Calendar.getInstance();
    calendar2.setTime(time2);
    calendar2.add(Calendar.DATE, 1);

    String someRandomTime = "01:00:00";
    Date d = new SimpleDateFormat("HH:mm:ss").parse(someRandomTime);
    Calendar calendar3 = Calendar.getInstance();
    calendar3.setTime(d);
    calendar3.add(Calendar.DATE, 1);

    Date x = calendar3.getTime();
    if (x.after(calendar1.getTime()) && x.before(calendar2.getTime())) {
        //checkes whether the current time is between 14:49:00 and 20:11:13.
        System.out.println(true);
    }
} catch (ParseException e) {
    e.printStackTrace();
}

相关文章