问题

我有以下的Long变量，其中保存着以秒为单位的时间戳，我想要将其转换为一个Date对象。

val seconds = 1341855763000
val date = Date(TimeUnit.SECONDS.toMillis(seconds))

输出的结果与我预期的相差甚远。我做错了什么？

实际输出：Wednesday, September 19, 44491 5:26:40 AM GMT+05:30
预期输出：Monday, July 9, 2012 11:12:43 AM GMT+05:30

val seconds = 1341855763000
val date = Date(TimeUnit.SECONDS.toMillis(seconds))

Actual: Wed Sep 19 05:26:40 GMT+05:30 44491
Expected: Monday July 9 11:12:43 GMT+05:30 2012

答案1

得分: 7

以下是翻译好的部分：

> 输出结果与我的预期相差甚远。我哪里做错了？

实际结果：Wed Sep 19 05:26:40 GMT+05:30 44491
预期结果：Monday July 9 11:12:43 GMT+05:30 2012

这个值已经是以 毫秒 为单位，通过使用 TimeUnit.SECONDS.toMillis(seconds)，你错误地将它乘以了 1000。

使用[现代日期时间 API][1]：

import java.time.Instant;

public class Main {
    public static void main(String[] args) {
        Instant instant = Instant.ofEpochMilli(1341855763000L);
        System.out.println(instant);
    }
}

输出结果：

2012-07-09T17:42:43Z

使用传统的 java.util.Date：

import java.util.Date;

public class Main {
    public static void main(String[] args) {
        System.out.println(new Date(1341855763000L));
    }
}

输出结果：

Mon Jul 09 18:42:43 BST 2012
我建议你从过时且容易出错的 `java.util` 日期时间 API 和 `SimpleDateFormat` 切换到[现代的](https://www.oracle.com/technical-resources/articles/java/jf14-date-time.html) `java.time` 日期时间 API 以及相应的格式化 API（包名为 `java.time.format`）。从**[教程：日期时间](https://docs.oracle.com/javase/tutorial/datetime/index.html)**中了解有关现代日期时间 API 的更多信息。

  [1]: https://www.oracle.com/technical-resources/articles/java/jf14-date-time.html

<details>
<summary>英文:</summary>

&gt; The output is way off than I expected. Where did I go wrong?

    Actual: Wed Sep 19 05:26:40 GMT+05:30 44491
    Expected: Monday July 9 11:12:43 GMT+05:30 2012

The value is already in `milliseconds` and by using `TimeUnit.SECONDS.toMillis(seconds)` you are wrongly multiplying it by `1000`.

## By using the [modern date-time API][1]:

    import java.time.Instant;
    
    public class Main {
    	public static void main(String[] args) {
    		Instant instant = Instant.ofEpochMilli(1341855763000L);
    		System.out.println(instant);
    	}
    }
**Output:**

    2012-07-09T17:42:43Z

## By using legacy `java.util.Date`:

    import java.util.Date;
    
    public class Main {
    	public static void main(String[] args) {
    		System.out.println(new Date(1341855763000L));
    	}
    }
**Output:**

    Mon Jul 09 18:42:43 BST 2012
I recommend you switch from the outdated and error-prone `java.util` date-time API and `SimpleDateFormat` to the [modern](https://www.oracle.com/technical-resources/articles/java/jf14-date-time.html) `java.time` date-time API and the corresponding formatting API (package, `java.time.format`). Learn more about the modern date-time API from **[Trail: Date Time](https://docs.oracle.com/javase/tutorial/datetime/index.html)**.

  [1]: https://www.oracle.com/technical-resources/articles/java/jf14-date-time.html

</details>



# 答案2
**得分**: 6

你所拥有的值不是以秒为单位，而是以毫秒为单位。移除“秒转毫秒”的转换。

val milliSeconds = 1341855763000
val date = Date(milliSeconds)

<details>
<summary>英文:</summary>

The value you have is not in seconds but in milliseconds. Remove the &quot;seconds to millis&quot; conversion.

val milliSeconds = 1341855763000
val date = Date(milliSeconds)

答案3

得分: 4

Your value : 1341855763000 is not in seconds, it is in milliseconds.
The current timestamp is :
new Date().getTime() =>
1598612990351
Same number of digits than :
1341855763000

If you multiply 1341855763000 by 1000 (as you say), it gives the year :
44491 after JC
Have a good day

答案4

得分: 3

我认为你的时间实际上是以毫秒为单位的。如果我使用这个网站将 1341855763000 进行转换，它会给我你期望的时间，还有这个：

fun main() {
  val millis = 1341855763000
  val date = Date(millis)
  println(date)
}

或者你也可以使用秒数：

fun main() {
  val seconds = 1341855763L
  val date = Date(TimeUnit.SECONDS.toMillis(seconds))
  println(date)
}

只需要将数值除以 1000。

fun main() {
  val millis = 1341855763000
  val date = Date(millis)
  println(date)
}

fun main() {
  val seconds = 1341855763L
  val date = Date(TimeUnit.SECONDS.toMillis(seconds))
  println(date)
}