java - 如何使用Java减去两个系统时间

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已关闭 4 年前。

我编写了一个Selenium脚本，我想计算该脚本的总执行时间。如何减去脚本开始执行和结束时系统返回的时间？

我正在使用Date和SimpleDateFormat Java 类来获取系统时间，然后对其进行格式化，但该方法似乎是错误的，我遵循的是返回总时间。

    String dateStart = "01/14/2012 23:05:49";
    String dateStop = "01/15/2012 23:07:00";

    SimpleDateFormat sdf = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss");

    Date d1 = null;
    Date d2 = null;

    try {
        d1 = sdf.parse(dateStart);
        d2 = sdf.parse(dateStop);

        //in milliseconds
        long diff = d2.getTime() - d1.getTime();

        long diffSeconds = diff / 1000 % 60;
        long diffMinutes = diff / (60 * 1000) % 60;
        long diffHours = diff / (60 * 60 * 1000) % 24;
        //long diffDays = diff / (24 * 60 * 60 * 1000);

        //System.out.print(diffDays + " days, ");
        System.out.print(diffHours + " hours, ");
        System.out.print(diffMinutes + " minutes, ");
        System.out.print(diffSeconds + " seconds.");

    } catch (Exception e) {
        e.printStackTrace();
    }

Answer 1

使用下面的代码:-

TimeUnit 枚举

以下表达式使用 TimeUnit枚举(Java 5 及更高版本)从纳秒转换为秒:

public static void main (String[] args) { 
long start = System.nanoTime(); 
//some try with nested loops 
long end = System.nanoTime(); 
long elapsedTime = end - start;

System.out.println("elapsed: " + elapsedTime + "nano seconds\n");

//convert to seconds 
TimeUnit seconds = new TimeUnit(); 
System.out.println("which is " + TimeUnit.NANOSECONDS.toSeconds(elapsedTime) + " seconds"); 
}}