Java 使用yyyy-MM-dd hh:MM:ss格式分析的日期存在意外差异

我已经运行了下面的java代码来获取时差

import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;
import java.util.TimeZone;


public class Test 
{
    public static SimpleDateFormat simpleDateFormat= new SimpleDateFormat("yyyy-MM-dd hh:mm:ss");
    public static Date date1,date2;
    public static long diff; 
    public static String TAG ="DateConversion";
    public static Calendar cal1,cal2;

    public static void main(String a[])
    {
        checkTimeDifference("2013-10-30 10:15:00", "2013-10-30 11:15:00");
        checkTimeDifference("2013-10-30 10:15:00", "2013-10-30 12:15:00");
        checkTimeDifference("2013-10-30 10:15:00", "2013-10-30 13:15:00");
    }

    public static  void checkTimeDifference(String strDate,String checkDate)
    {
        try 
        {
            simpleDateFormat.setTimeZone(TimeZone.getTimeZone("GMT"));
            date1    = simpleDateFormat.parse(strDate);
            date2    = simpleDateFormat.parse(checkDate);

            //in milliseconds
            diff = date2.getTime() - date1.getTime();
            System.out.println("Difference : "+diff);
            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.println(diffDays     + " days, ");
            System.out.println(diffHours    + " hours, ");
            System.out.println(diffMinutes+ " minutes, ");
            System.out.println(diffSeconds+ " seconds.");
        }
        catch (Exception e) 
        {
            System.out.println(""+e);
        }
    }
}

上述程序的输出为

Difference : 3600000
0 days, 
1 hours, 
0 minutes, 
0 seconds.
Difference : -36000000
0 days, 
-10 hours, 
0 minutes, 
0 seconds.
Difference : 10800000
0 days, 
3 hours, 
0 minutes, 
0 seconds.

执行checkTimeDifference（“2013-10-30 10:15:00”、“2013-10-30 12:15:00”）时，其返回值减去值；“

为什么它的返回值为负值？如何解决它？

这就是问题所在：

new SimpleDateFormat("yyyy-MM-dd hh:mm:ss")

这里的

hh

意思是“一天中的12小时”，所以12表示午夜，除非有什么东西表明它是指下午12点。您的值13只起作用，因为解析器处于宽松模式。你想要：

new SimpleDateFormat("yyyy-MM-dd HH:mm:ss")

---

我还强烈建议您无论如何都要使用它来完成此任务，因为它会使任务变得更简单。

将

SimpleDateFormat

模式更改为

yyyy-MM-dd HH:MM:ss

可以解决问题

这是因为在

yyyy-MM-dd hh:MM:ss

的情况下，

12

被评估为

0