检查LocalDateTime是否在一个时间范围内 [英] Checking if LocalDateTime falls within a time range

问题描述

我有一个时间A应该在90分钟的时间范围B（之前和之后）。

I have a time A which should fall within 90 minutes range of timeB (before and after).

示例：如果timeB是下午4:00，则时间A应该在下午2:30（-90）到下午5:30（+90）之间

Example: if timeB is 4:00 pm , time A should be between 2:30pm (-90) to 5:30pm (+90)

试过以下内容：

if(timeA.isAfter(timeB.minusMinutes(90)) || timeA.isBefore(timeB.plusMinutes(90))) {
    return isInRange;   
}

你能帮助我解决这里的逻辑错误吗？

Can you help me whats wrong with the logic here?

推荐答案

as @ JB Nizet在评论中说，您使用的是 OR 运算符（ || ）。$
所以你要测试的是 A是否在B-90之后 或 A在B + 90之前。如果只满足其中一个条件，则返回 true 。

As @JB Nizet said in the comments, you're using the OR operator (||).
So you're testing if A is after B - 90 OR A is before B + 90. If only one of the conditions is satisfied, it returns true.

检查 在此范围内，必须满足这两个条件，因此您必须使用 AND 运算符（&& ）：

To check if A is in the range, both conditions must be satisfied, so you must use the AND operator (&&):

if (timeA.isAfter(timeB.minusMinutes(90)) && timeA.isBefore(timeB.plusMinutes(90))) {
    return isInRange;   
}

---

但上面的代码没有返回 true 如果 A 在<$ c $之前或之后90分钟完全 C> B 。如果您希望它返回 true ，当差异也是90分钟时，您必须更改条件以检查：

---

But the code above doesn't return true if A is exactly 90 minutes before or after B. If you want it to return true when the difference is also exactly 90 minutes, you must change the condition to check this:

// lower and upper limits
LocalDateTime lower = timeB.minusMinutes(90);
LocalDateTime upper = timeB.plusMinutes(90);
// also test if A is exactly 90 minutes before or after B
if ((timeA.isAfter(lower) || timeA.equals(lower)) && (timeA.isBefore(upper) || timeA.equals(upper))) {
    return isInRange;
}

---

另一种选择是使用一个 java.time.temporal.ChronoUnit 来区分 A 和 B 以分钟为单位，并检查其价值：

---

Another alternative is to use a java.time.temporal.ChronoUnit to get the difference between A and B in minutes, and check its value:

// get the difference in minutes
long diff = Math.abs(ChronoUnit.MINUTES.between(timeA, timeB));
if (diff <= 90) {
    return isInRange;
}

我用 Math.abs 因为如果 A 在 B 之后，差异可能是负数（因此它被调整为正数）。然后我检查差异是否小于（或等于）90分钟。您可以将其更改为 if（diff< 90）如果您要排除等于90分钟的情况。

I used Math.abs because the difference can be negative if A is after B (so it's adjusted to be a positive number). Then I check if the difference is less than (or equal) to 90 minutes. You can change it to if (diff < 90) if you want to exclude the "equals to 90 minutes" case.

这些方法之间存在差异。

There's a difference between the approaches.

ChronoUnit 改变差异。例如如果 A 在 B 之后90分59秒，则差异将四舍五入为90分钟且 if（diff< = 90）将 true ，同时使用 isBefore 和等于将返回 false 。

ChronoUnit rounds the difference. e.g. If A is 90 minutes and 59 seconds after B, the difference will be rounded to 90 minutes and if (diff <= 90) will be true, while using isBefore and equals will return false.

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